If you feel that you would not be confident if questions similar to those below appeared on your exam, we recommend booking one of our one-to-one grinds!
Exercise Set 6B
These questions are considered advanced Ordinary Level to beginner Higher Level.
Question 1
An object is released from rest from a height of \(5\) metres.
Find the speed that the object strikes the ground:
(a) using a SUVAT equation
(b) using the principle of conservation of energy
Answer
(a) \(\sqrt{98}\mbox{ m/s}\)
(b) \(\sqrt{98}\mbox{ m/s}\)
Solution
(a)
\begin{align}u=0 &&a=-9.8 && s=-5&&v=?\end{align}
\[v^2 = u^2 + 2as\]
\begin{align}\downarrow\end{align}
\begin{align}v^2 &= 0^2 + 2(-9.8)(-5)\\&=98\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v=\sqrt{98}\mbox{ m/s}\end{align}
(b)
\[mgh = \frac{1}{2}mv^2\]
\begin{align}\downarrow\end{align}
\begin{align}v&=\sqrt{2gh}\\&=\sqrt{2(9.8)(5)}\\&=\sqrt{98}\mbox{ m/s}\end{align}
Walkthrough
(a)
Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!
We are given the following coefficients
\begin{align}u=0 &&a=-9.8 && s=-5\end{align}
and we are looking for \(v\).
\[\,\]
Step 2: Choose the most suitable SUVAT equation based on Step 1.
As we are not told \(t\) nor do we want to find \(t\), we should choose the SUVAT equation which does not have \(t\) in it, namely
\[v^2 = u^2 + 2as\]
\[\,\]
Step 3: Insert the quantities into the equation.
Subbing in our known values, we obtain
\begin{align}v^2 &= 0^2 + 2(-9.8)(-5)\\&=98\end{align}
Therefore, the speed of the object when it strikes the ground is \(\sqrt{98}\mbox{ m/s}\).
(b) If we assume that the system is isolated, i.e. that \(E_i=E_f\), we obtain
\[mgh = \frac{1}{2}mv^2\]
(as the initial energy when released is entirely potential energy and final energy when it reaches the ground is entirely kinetic energy).
Cancelling the masses and rearranging, we obtain
\begin{align}v&=\sqrt{2gh}\\&=\sqrt{2(9.8)(5)}\\&=\sqrt{98}\mbox{ m/s}\end{align}
Question 2
A pendulum consists of a light inextensible string of length \(60\mbox{ cm}\) with one end connected to a ceiling and the other end connected to an object of mass \(2\mbox{ kg}\).
The object is initially held as shown below.
When released, what is the maximum kinetic energy of the object during its subsequent motion?
Answer
\(3.53\mbox{ J}\)
Solution
Initially, the object is \(0.6 \cos 45^{\circ}\approx 0.42\) metres from the ceiling.
Therefore, initially, the object is \(0.6-0.42=0.18\mbox{ m}\) higher from the ground than it is when at its lowest point.
Initial Energy
\begin{align}U_{i}&=mgh\\&=(2)(9.8)(0.18)\\&\approx 3.53 \mbox{ J}\end{align}
and
\begin{align}K_i = 0 \mbox{ J}\end{align}
Final Energy
\begin{align}U_{f}&=0 \mbox{ J}\end{align}
and
\(K_f\) is what we’re looking for
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}3.53+0=0+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}K_f=3.53 \mbox{ J} \end{align}
Walkthrough
The kinetic energy of the object has a maximum value when it as its lowest point (as its has its greatest speed at this point).
To find this value, let us therefore apply the principle of conservation of energy.
Let’s refer to the height of the object when its at its lowest point as our reference location \(h=0\). How much higher than that is the object initially?
Well, at its lowest point, the object is \(0.6\mbox{ m}\) from the ceiling (as that’s the length of the string).
Initially, the object is instead \(0.6 \cos 45^{\circ}\approx 0.42\) metres from the ceiling.
Therefore, initially, the object is \(0.6-0.42=0.18\mbox{ m}\) higher from the ground than it is when at its lowest point.
Hence, the object’s initial potential energy is
\begin{align}U_{i}&=mgh\\&=(2)(9.8)(0.18)\\&\approx 3.53 \mbox{ J}\end{align}
As the object is initially at rest, the object has an initial kinetic energy of
\begin{align}K_i &= \frac{1}{2}mu^2\\&=\frac{1}{2}(2)(0^2)\\&=0\ \mbox{ J}\end{align}
As we have set \(h=0\) as the height of the object at its lowest point, its potential energy at that point is
\begin{align}U_f&=mgh\\&=(2)(9.8)(0)\\&=0 \mbox{ J}\end{align}
Finally, the kinetic energy at the lowest points is the final kinetic energy \(K_f\) that we are looking to find.
Therefore, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}3.53+0=0+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}K_f=3.53 \mbox{ J} \end{align}
Question 3
A football is placed on a smooth surface which is inclined at an angle of \(30^{\circ}\) relative to the horizontal.
What is the object’s speed after it has travelled a distance of \(10\mbox{ m}\)?
Answer
\(9.90\mbox{ m/s}\)
Solution
The object’s height is initially \(10 \sin 30^{\circ} = 5\) metres.
Initial Energy
\begin{align}U_{i}&=mgh\\&=m(9.8)(5)\\&=49m\end{align}
and
\begin{align}K_i =0 \mbox{ J}\end{align}
Final Energy
\begin{align}U_f=0 \mbox{ J}\end{align}
and
\begin{align}K_f &= \frac{1}{2}mv^2\end{align}
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}49m+0=0+\frac{1}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}49=\frac{1}{2}v^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v&=\pm \sqrt{2(49)}\\&\approx \pm 9.90 \mbox{ m/s}\end{align}
Hence, the object’s speed after it has travelled a distance of \(10\mbox{ m}\) down the incline is \(9.90\mbox{ m/s}\).
Walkthrough
This question can be answered in one of two ways – we can either use SUVAT equations with rotated axes or the principle of conservation of energy. We shall randomly choose the latter approach.
Let’s refer to the height of the object when its at its lowest point as our reference location \(h=0\), i.e. \(h=0\) corresponds to height of the object after it has travelled \(10\mbox{ m}\) down the incline. How much higher than that is the object initially?
Well, according to the below, the object’s height is initially \(10 \sin 30^{\circ} = 5\) metres.
Hence, the object’s initial potential energy is
\begin{align}U_{i}&=mgh\\&=m(9.8)(5)\\&=49m\end{align}
As the object is initially at rest, the object has an initial kinetic energy of
\begin{align}K_i &= \frac{1}{2}mv^2\\&=\frac{1}{2}m(0^2)\\&=0 \mbox{ J}\end{align}
As we have set \(h=0\) to be the height of the object at its lowest point, its potential energy at that point is
\begin{align}U_f&=mgh\\&=m(9.8)(0)\\&=0 \mbox{ J}\end{align}
Finally, if we set the object’s speed at its lowest point to \(v\), its final kinetic energy is
\begin{align}K_f &= \frac{1}{2}mv^2\end{align}
Therefore, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}49m+0=0+\frac{1}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}49=\frac{1}{2}v^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v&=\pm \sqrt{2(49)}\\&\approx \pm 9.90\end{align}
Hence, the object’s speed after it has travelled a distance of \(10\mbox{ m}\) down the incline is \(9.90\mbox{ m/s}\).
Question 4
An object of mass \(m\) is released from a height \(h\) above the ground.
Show that the work done by gravity when it strikes the grounds is \(mgh\)
(a) using \(\vec{W} = \vec{F} \cdot \vec{s}\)
(b) using the Work-Energy theorem and the principle of conservation of energy
Answer
(a) The answer is already in the question!
(b) The answer is already in the question!
Solution
(a) \(\vec{W} = -mg\vec{j}\) and \(\vec{s}=-h\vec{j}\). Hence
\begin{align}W&=\vec{W}\cdot \vec{s}\\&=(-mg\vec{j})\cdot(-h\vec{j})\\&=(-mg)(-h)(\vec{j}\cdot\vec{j})\\&=mgh\end{align}
as required.
(b)
Initial Energy
\begin{align}U_i=mgh\end{align}
and
\begin{align}K_i =0\mbox{ J} \end{align}
Final Energy
\begin{align}U_f=0 \mbox{ J}\end{align}
and
\begin{align}K_f = \frac{1}{2}mv^2 \end{align}
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i=U_f+K_f\end{align}
\begin{align}\downarrow\end{align}
\begin{align}mgh+0=0+\frac{1}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}mgh=\frac{1}{2}mv^2\end{align}
According to the work-energy theorem, the work done by gravity is
\begin{align}W&=K_f-K_i\\&=\frac{1}{2}mv^2-0\\&=\frac{1}{2}mv^2\end{align}
Using our relation obtained using conservation of energy above, we can instead rewrite this as
\begin{align}W=mgh\end{align}
as required.
Walkthrough
(a) The gravitational force on the object, i.e. the object’s weight, has a magnitude of \(mg\) and points in the negative \(y\) direction, i.e. \(\vec{W} = -mg\vec{j}\).
The displacement vector has a magnitude of \(h\) and is also in the negative \(y\) direction, i.e. \(\vec{s}=-h\vec{j}\).
Hence, the work done by gravity is
\begin{align}W&=\vec{W}\cdot \vec{s}\\&=(-mg\vec{j})\cdot(-h\vec{j})\\&=(-mg)(-h)(\vec{j}\cdot\vec{j})\\&=mgh\end{align}
as required.
(b) Let’s set the reference location \(h=0\) to be the ground.
As the object is initially at a height \(h\), its initial potential energy is
\begin{align}U_i=mgh\end{align}
As the object is released from rest, its initial kinetic energy is
\begin{align}K_i &= \frac{1}{2}mu^2\\&=\frac{1}{2}m(0^2)\\&=0 \end{align}
As the object is finally at the reference location \(h=0\), its final potential energy is
\begin{align}U_f&=mgh\\&=mg(0)\\&=0\end{align}
Let \(v\) be the object’s kinetic energy when it strikes the ground. Its final kinetic energy is therefore
\begin{align}K_f = \frac{1}{2}mv^2 \end{align}
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i=U_f+K_f\end{align}
\begin{align}\downarrow\end{align}
\begin{align}mgh+0=0+\frac{1}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}mgh=\frac{1}{2}mv^2\end{align}
According to the work-energy theorem, the work done by gravity is
\begin{align}W&=K_f-K_i\\&=\frac{1}{2}mv^2-0\\&=\frac{1}{2}mv^2\end{align}
Using our relation obtained using conservation of energy above, we can instead rewrite this as
\begin{align}W=mgh\end{align}
as required.
Question 5
An object is hanging vertically from the end of a string of length \(2\) metres.
The object is given a push horizontally, resulting in an initial speed of \(10\mbox{ m/s}\).
(a) What is the object’s speed when it vertically above its starting position?
(b) If we wish for the object to instead only just make it to this position, i.e. for it to come to rest at this position, what must the object’s speed instead be when it is pushed?
Answer
(a) \(4.65\mbox{ m/s}\)
(b) \(8.85\mbox{ m/s}\)
Solution
(a)
Initial Energy
\begin{align}U_{i}=0\mbox{ J}\end{align}
and
\begin{align}K_i &= \frac{1}{2}mu^2\\&=\frac{1}{2}m(10^2)\\&=50m\end{align}
Final Energy
\begin{align}U_f&=mgh\\&=m(9.8)(4)\\&=39.2m\end{align}
and
\begin{align}K_f &= \frac{1}{2}mv^2\end{align}
Therefore, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}0+50m=39.2m+\frac{1}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}v^2=10.8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v&=\pm \sqrt{2(10.8)}\\&\approx \pm 4.65\end{align}
Therefore, the object’s speed when it is vertically above its starting position is \(4.65\mbox{ m/s}\).
(b)
Initial Energy
\begin{align}U_{i}=0\mbox{ J}\end{align}
and
\begin{align}K_i &= \frac{1}{2}mu^2\end{align}
Final Energy
\begin{align}U_f&=mgh\\&=m(9.8)(4)\\&=39.2m\end{align}
and
\begin{align}K_f =0\mbox{ J}\end{align}
Therefore, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}0+\frac{1}{2}mu^2=39.2m0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}u^2=39.2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}u&=\pm \sqrt{2(39.2)}\\&\approx \pm 8.85\end{align}
Therefore, if the object is to come to rest when it is vertically above its initial position, the object’s speed when pushed must be \(8.85\mbox{ m/s}\).
Walkthrough
(a) Let’s set the reference location \(h=0\) to the height of the object when it is pushed.
Hence, the object’s initial potential energy is
\begin{align}U_{i}&=mgh\\&=m(9.8)(0)\\&=0\mbox{ J}\end{align}
As the object’s speed at this position is \(10\) m/s, the object has an initial kinetic energy of
\begin{align}K_i &= \frac{1}{2}mu^2\\&=\frac{1}{2}m(10^2)\\&=50m\end{align}
The object will be vertically above this position after it has travelled in a semi-circular arc of radius \(2\) metres. The height of the object in this position is therefore one diameter larger than its initial height, i.e. \(h=2\times 2=4\) metres.
Hence, the object’s final potential energy is
\begin{align}U_f&=mgh\\&=m(9.8)(4)\\&=39.2m\end{align}
Finally, if we set the object’s speed at this final position to \(v\), its final kinetic energy is
\begin{align}K_f &= \frac{1}{2}mv^2\end{align}
Therefore, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}0+50m=39.2m+\frac{1}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}v^2=10.8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v&=\pm \sqrt{2(10.8)}\\&\approx \pm 4.65\end{align}
Therefore, the object’s speed when it is vertically above its starting position is \(4.65\mbox{ m/s}\).
(b) Again, let’s set the reference location \(h=0\) to the height of the object when it is pushed.
Hence, the object’s initial potential energy is
\begin{align}U_{i}&=mgh\\&=m(9.8)(0)\\&=0\mbox{ J}\end{align}
If we set the object’s speed when it is pushed as \(u\), the object has an initial kinetic energy of
\begin{align}K_i &= \frac{1}{2}mu^2\end{align}
Again, when vertically above its starting position, the height of the object is one diameter larger than its initial height, i.e. \(h=2\times 2=4\) metres.
Hence, the object’s final potential energy is
\begin{align}U_f&=mgh\\&=m(9.8)(4)\\&=39.2m\end{align}
Finally, if the object comes to rest at the final position, its final kinetic energy is
\begin{align}K_f &= \frac{1}{2}mv^2\\&=\frac{1}{2}m(0^2)\\&=0\mbox{ J}\end{align}
Therefore, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}0+\frac{1}{2}mu^2=39.2m0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}u^2=39.2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}u&=\pm \sqrt{2(39.2)}\\&\approx \pm 8.85\end{align}
Therefore, if the object is to come to rest when it is vertically above its initial position, the object’s speed when pushed must be \(8.85\mbox{ m/s}\).
Question 6
Object \(A\) of mass \(m\) is placed on one end of a light seesaw, as shown below.
Object \(B\) of mass \(3m\) is then placed on the other end of the seesaw.
Find the common speed of the system when object \(B\) reaches the ground.
Answer
\(3.13\mbox{ m/s}\)
Solution
Object A
Initial Energy
\begin{align}U_{Ai}=0\mbox{ J}\end{align}
and
\begin{align}K_{Ai}=0\mbox{ J}\end{align}
Final Energy
\begin{align}U_{Af}&=m_Agh\\&=m(9.8)(1)\\&=9.8m\end{align}
and
\begin{align}K_{Af}&=\frac{1}{2}mv^2\end{align}
Object B
Initial Energy
\begin{align}U_{Bi}&=m_Bgh\\&=(3m)(9.8)(1)\\&=29.4m\end{align}
and
\begin{align}K_{Bi}=0\mbox{ J}\end{align}
Final Energy
\begin{align}U_{Bf}=0\mbox{ J}\end{align}
and
\begin{align}K_{Bf}&=\frac{1}{2}m_Bv^2\\&=\frac{1}{2}(3m)v^2\\&=\frac{3}{2}mv^2\end{align}
Therefore, according to the principle of conservation of energy
\begin{align}U_{Ai}+K_{Ai}+U_{Bi}+K_{Bi}=U_{Af}+K_{Af}+U_{Bf}+K_{Bf}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0+0+29.4m+0=9.8m+\frac{1}{2}mv^2+0+\frac{3}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}19.6m=2mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v^2=9.8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v&\approx \pm 3.13 \mbox{ m/s}\end{align}
Therefore, when object \(B\) reaches the ground, the common speed of the system is \(3.13\mbox{ m/s}\).
Walkthrough
As object \(B\) is heavier, that object will cause the right hand side of the see-saw to go down while the left hand side goes up. Let the ground be the reference location \(h=0\).
As object \(A\) is initially at height of \(0\) metres, it has an initial potential energy of
\begin{align}U_{Ai}&=m_Agh\\&=m(9.8)(0)\\&=0\mbox{ J}\end{align}
As object \(A\) is initially at rest, it has an initial kinetic energy of
\begin{align}K_{Ai}&=\frac{1}{2}m_Au_{Ai}^2\\&=\frac{1}{2}m(0^2)\\&=0\mbox{ J}\end{align}
As object \(B\) is initially at a height of \(1\) metre, it has an initial potential energy of
\begin{align}U_{Bi}&=m_Bgh\\&=(3m)(9.8)(1)\\&=29.4m\end{align}
As object \(B\) is also initially at rest, it has an initial kinetic energy of
\begin{align}K_{Bi}&=\frac{1}{2}m_Bu_{Bi}^2\\&=\frac{1}{2}(3m)(0^2)\\&=0\mbox{ J}\end{align}
As object \(A\) is finally at a height of \(1\) metre, it has a final potential energy of
\begin{align}U_{Af}&=m_Agh\\&=m(9.8)(1)\\&=9.8m\end{align}
If we label the final common speed of the system as \(v\), then object \(A\) has a final kinetic energy of
\begin{align}K_{Af}&=\frac{1}{2}mv^2\end{align}
As object \(B\) is finally at a height of \(0\) metres, it has a final potential energy of
\begin{align}U_{Bf}&=m_Bgh\\&=(3m)(9.8)(0)\\&=0\mbox{ J}\end{align}
Object \(B\) has a final kinetic energy of
\begin{align}K_{Bf}&=\frac{1}{2}m_Bv^2\\&=\frac{1}{2}(3m)v^2\\&=\frac{3}{2}mv^2\end{align}
Therefore, according to the principle of conservation of energy
\begin{align}U_{Ai}+K_{Ai}+U_{Bi}+K_{Bi}=U_{Af}+K_{Af}+U_{Bf}+K_{Bf}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0+0+29.4m+0=9.8m+\frac{1}{2}mv^2+0+\frac{3}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}19.6m=2mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v^2=9.8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v&\approx \pm 3.13 \mbox{ m/s}\end{align}
Therefore, when object \(B\) reaches the ground, the common speed of the system is \(3.13\mbox{ m/s}\).
Question 7
A football of mass \(400\mbox{ g}\) is placed on a surface which is inclined at an angle of \(30^{\circ}\) relative to the horizontal.
The object’s speed after it has travelled a distance of \(5\mbox{ m}\) is \(12\mbox{ m/s}\).
How much energy has the football lost during this motion?
Answer
\(4.8\mbox{ J}\)
Solution
The object’s height is initially \(5 \sin 30^{\circ} = 2.5\) metres.
Initial Energy
\begin{align}U_{i}&=mgh\\&=(0.4)(9.8)(2.5)\\&=9.8 \mbox{ J}\end{align}
and
\begin{align}K_i =0 \mbox{ J}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}E_i&=U_i+K_i\\&=9.8+0\\&=9.8 \mbox{ J}\end{align}
Final Energy
\begin{align}U_f=0 \mbox{ J}\end{align}
and
\begin{align}K_f &= \frac{1}{2}mv^2\\&=\frac{1}{2}(0.4)(5^2)\\&=5 \mbox{ J}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}E_f&=U_f+K_f\\&=0+5\\&=5 \mbox{ J}\end{align}
Hence, the amount of energy lost during this motion was
\begin{align}E_i-E_f &= 9.8-5\\&=4.8 \mbox{ J}\end{align}
Walkthrough
We wish to find the difference between the initial energy \(E_i=U_i+K_i\) and the final energy \(E_f=U_f+K_f\) of the football.
Let’s refer to the height of the object when its at its lowest point as our reference location \(h=0\), i.e. \(h=0\) corresponds to height of the object after it has travelled \(5\) m down the incline. How much higher than that is the object initially?
Well, according to the below, the object’s height is initially \(5 \sin 30^{\circ} = 2.5\) metres.
Hence, the object’s initial potential energy is
\begin{align}U_{i}&=mgh\\&=(0.4)(9.8)(2.5)\\&=9.8 \mbox{ J}\end{align}
As the object is initially at rest, the object has an initial kinetic energy of
\begin{align}K_i &= \frac{1}{2}mv^2\\&=\frac{1}{2}(0.4)(0^2)\\&=0 \mbox{ J}\end{align}
Therefore, the object’s initial total energy is
\begin{align}E_i&=U_i+K_i\\&=9.8+0\\&=9.8 \mbox{ J}\end{align}
As we have set \(h=0\) to be the height of the object at its lowest point, its potential energy at that point is
\begin{align}U_f&=mgh\\&=(0.4)(9.8)(0)\\&=0 \mbox{ J}\end{align}
Finally, as the object’s speed at this lowest point is \(5\) m/s, its final kinetic energy is
\begin{align}K_f &= \frac{1}{2}mv^2\\&=\frac{1}{2}(0.4)(5^2)\\&=5 \mbox{ J}\end{align}
Therefore, the object’s final total energy is
\begin{align}E_f&=U_f+K_f\\&=0+5\\&=5 \mbox{ J}\end{align}
Hence, the amount of energy lost during this motion was
\begin{align}E_i-E_f &= 9.8-5\\&=4.8 \mbox{ J}\end{align}
As energy is lost from the system, the system must not be closed. Energy could have been, for example, transferred from the system to the surface if the surface is rough.
Question 8
An object is thrown vertically upwards from a height of \(2\) metres with a speed of \(12\mbox{ m/s}\).
Find the maximum height of the object
(a) using a SUVAT equation
(b) using the principle of conservation of energy
The object strikes the ground with a speed \(v\).
(c) Find the value of \(v\) using any preferred method.
Answer
(a) \(9.35\mbox{ m}\)
(b) \(9.35\mbox{ m}\)
(c) \(13.54\mbox{ m/s}\)
Solution
\begin{align} u = 12 && a=-9.8 && v=0 && s=\mathord{?} \end{align}
\begin{align}\downarrow\end{align}
\[v^2=u^2+2as\]
\begin{align}\downarrow\end{align}
\begin{align}0^2=12^2+2(-9.8)s\end{align}
\begin{align}\downarrow\end{align}
\begin{align}s&=-\frac{12^2}{2(-9.8)}\\&\approx 7.35 \mbox{ m}\end{align}
Therefore, as the object is initially \(2\) metres from the ground, its maximum height is \(9.35\mbox{ m}\).
(b)
Initial Energy
\begin{align}U_i&=mgh\\&=m(9.8)(2)\\&=19.6m\end{align}
and
\begin{align}K_i &= \frac{1}{2}mu^2\\&=\frac{1}{2}m(12^2)\\&=72m \end{align}
Final Energy
\begin{align}U_f&=mgh\\&=9.8mh\end{align}
and
\begin{align}K_f =0\mbox{ J} \end{align}
(b)
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i=U_f+K_f\end{align}
\begin{align}\downarrow\end{align}
\begin{align}19.6m+72m=9.8mh+0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}91.6m=9.8mh\end{align}
\begin{align}\downarrow\end{align}
\begin{align}91.6=9.8h\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h&=\frac{91.6}{9.8}\\&=9.35 \mbox{ m}\end{align}
(c)
Initial Energy
\begin{align}U_i&=mgh\\&=m(9.8)(2)\\&=19.6m\end{align}
and
\begin{align}K_i &= \frac{1}{2}mu^2\\&=\frac{1}{2}m(12^2)\\&=72m \end{align}
Final Energy
\begin{align}U_f=0\mbox{ J}\end{align}
and
\begin{align}K_f &= \frac{1}{2}mv^2 \end{align}
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i=U_f+K_f\end{align}
\begin{align}\downarrow\end{align}
\begin{align}19.6m+72m=0+\frac{1}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}91.6m=\frac{1}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}91.6=\frac{1}{2}v^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v&=\pm\sqrt{2(91.6)}\\&\approx \pm 13.54 \mbox{ m/s}\end{align}
Therefore, the object strikes the ground with a speed of \(13.54\mbox{ m/s}\).
Walkthrough
(a) Let’s set the initial height of the object as our reference location \(s=0\). At the maximum height, the speed of the object is zero.
Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!
We are given the following coefficients
\begin{align} u = 12 && a=-9.8 && v=0 \end{align}
and we are looking for \(s\).
\[\,\]
Step 3: Choose the most suitable SUVAT equation based on Step 2
As we are not told \(t\) nor do we want to find \(t\), we should choose the SUVAT equation which does not have \(t\) in it, namely
\[v^2=u^2+2as\]
\[\,\]
Step 4: Insert the quantities into the equation
Subbing in our known values, we obtain
\begin{align}0^2=12^2+2(-9.8)s\end{align}
Rearranging for \(s\), this gives
\begin{align}s&=-\frac{12^2}{2(-9.8)}\\&\approx 7.35 \mbox{ m}\end{align}
Therefore, as the object is initially \(2\) metres from the ground, its maximum height is \(9.35\mbox{ m}\).
(b) Let’s set the ground as our reference location \(h=0\).
As the object is initially at a height \(h=2\) metres, its initial potential energy is
\begin{align}U_i&=mgh\\&=m(9.8)(2)\\&=19.6m\end{align}
As the object has an initial speed of \(12\mbox{ m/s}\), its initial kinetic energy is
\begin{align}K_i &= \frac{1}{2}mu^2\\&=\frac{1}{2}m(12^2)\\&=72m \end{align}
Let us set the final height of the object as \(h\). Therefore, its final potential energy is
\begin{align}U_f&=mgh\\&=9.8mh\end{align}
At its maximum height, the object’s speed is zero. Therefore, its final kinetic energy is
\begin{align}K_f &= \frac{1}{2}mv^2\\&=\frac{1}{2}m(0^2)\\&=0\mbox{ J} \end{align}
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i=U_f+K_f\end{align}
\begin{align}\downarrow\end{align}
\begin{align}19.6m+72m=9.8mh+0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}91.6m=9.8mh\end{align}
Each term contains an \(m\) which can be cancelled, giving
\begin{align}91.6=9.8h\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h&=\frac{91.6}{9.8}\\&=9.35 \mbox{ m}\end{align}
(c) We can find this speed using either of our two approaches above. We shall randomly choose the conservation of energy approach.
Let us set the ground as our reference location \(h=0\).
As the object is initially at a height \(h=2\) metres, its initial potential energy is
\begin{align}U_i&=mgh\\&=m(9.8)(2)\\&=19.6m\end{align}
As the object’s initial speed is \(12\mbox{ m/s}\), its initial kinetic energy is
\begin{align}K_i &= \frac{1}{2}mu^2\\&=\frac{1}{2}m(12^2)\\&=72m \end{align}
As the object is finally at the reference location \(h=0\), its final potential energy is
\begin{align}U_f&=mgh\\&=m(9.8)(0)\\&=0\mbox{ J}\end{align}
Setting \(v\) as the object’s speed when it strikes the ground, the object’s final kinetic energy is
\begin{align}K_f &= \frac{1}{2}mv^2 \end{align}
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i=U_f+K_f\end{align}
\begin{align}\downarrow\end{align}
\begin{align}19.6m+72m=0+\frac{1}{2}mv^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}91.6m=\frac{1}{2}mv^2\end{align}
Each term contains an \(m\) which can be cancelled, giving
\begin{align}91.6=\frac{1}{2}v^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}v&=\pm\sqrt{2(91.6)}\\&\approx \pm 13.54 \mbox{ m/s}\end{align}
Therefore, the object strikes the ground with a speed of \(13.54\mbox{ m/s}\).
Question 9
A pendulum consists of a light inextensible string of length \(40\mbox{ cm}\) with one end connected to a ceiling and the other end connected to an object of mass \(4\mbox{ kg}\).
The object is initially held as shown below.
When released, the greatest speed of the object during its subsequent motion is \(2\mbox{ m/s}\).
What is value of \(\theta\)?
Answer
\(60.66^{\circ}\)
Solution
Initially, the object is \(0.4 \cos \theta\) metres from the ceiling.
Therefore, initially, the object is \(0.4-0.4 \cos\theta\) metres higher from the ground than it is when at is lowest point.
Initial Energy
\begin{align}U_{i}&=mgh\\&=(4)(9.8)(0.4-0.4\cos \theta)\\&=15.68-15.68 \cos \theta\end{align}
and
\begin{align}K_i =0\ \mbox{ J}\end{align}
Final Energy
\begin{align}U_f=0 \mbox{ J}\end{align}
and
\begin{align}K_f &= \frac{1}{2}mv^2\\&=\frac{1}{2}(4)(2^2)\\&= 8 \mbox{ J}\end{align}
Therefore, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}(15.68-15.68\cos \theta) + 0 = 0+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos \theta &= \frac{15.68-8}{15.68}\\&\approx 0.49\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta &= \cos^{-1} (0.49)\\&\approx 60.66^{\circ}\end{align}
Walkthrough
To find this angle, let us apply the principle of conservation of energy.
Let’s refer to the height of the object when its at its lowest point as our reference location \(h=0\), i.e. \(h=0\) corresponds to where the object has it greatest speed. How much higher than that is the object initially?
Well, at its lowest point, the object is \(0.4\mbox{ m}\) from the ceiling (as that’s the length of the string).
Initially, the object is instead \(0.4 \cos \theta\) metres from the ceiling.
Therefore, initially, the object is \(0.4-0.4 \cos\theta\) metres higher from the ground than it is when at is lowest point.
Hence, the object’s initial potential energy is
\begin{align}U_{i}&=mgh\\&=(4)(9.8)(0.4-0.4\cos \theta)\\&=15.68-15.68 \cos \theta\end{align}
As the object is initially at rest, the object has an initial kinetic energy of
\begin{align}K_i &= \frac{1}{2}mv^2\\&=\frac{1}{2}(4)(0^2)\\&=0\ \mbox{ J}\end{align}
As we have set \(h=0\) as the height of the object at its lowest point, its potential energy at that point is
\begin{align}U_f&=mgh\\&=(4)(9.8)(0)\\&=0 \mbox{ J}\end{align}
Finally, as the object is moving with a speed of \(2\mbox{ m/s}\) at its lowest point, its final kinetic energy is
\begin{align}K_f &= \frac{1}{2}mv^2\\&=\frac{1}{2}(4)(2^2)\\&= 8 \mbox{ J}\end{align}
Therefore, according to the principle of conservation of energy
\begin{align}U_i+K_i &=U_f+K_f \end{align}
\begin{align}\downarrow\end{align}
\begin{align}(15.68-15.68\cos \theta) + 0 = 0+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos \theta &= \frac{15.68-8}{15.68}\\&\approx 0.49\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta &= \cos^{-1} (0.49)\\&\approx 60.66^{\circ}\end{align}
Question 10
An object of mass \(m\) is thrown vertically upwards with a velocity \(\vec{u}\).
Using the principle of conservation of energy, show that the maximum height reached is
\begin{align}h = \frac{u^2}{2g}\end{align}
Answer
The answer is already in the question!
Solution
Initial Energy
\begin{align}U_i=0\mbox{ J}\end{align}
and
\begin{align}K_i = \frac{1}{2}mu^2 \end{align}
Final Energy
\begin{align}U_f&=mgh\end{align}
and
\begin{align}K_f =0\mbox{ J} \end{align}
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i=U_f+K_f\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0+\frac{1}{2}mu^2=mgh+0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}u^2=gh\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h=\frac{u^2}{2g}\end{align}
as required.
Walkthrough
Let’s set the reference location \(h=0\) to be the ground.
As the object is initially on the ground, its initial potential energy is
\begin{align}U_i&=mgh\\&=mg(0)\\&=0\mbox{ J}\end{align}
As the object is thrown with a speed \(u\), its initial kinetic energy is
\begin{align}K_i = \frac{1}{2}mu^2 \end{align}
As the object’s maximum height is \(h\), its final potential energy is
\begin{align}U_f&=mgh\end{align}
At the maximum height, the object’s speed is zero. Therefore, its final kinetic energy is
\begin{align}K_f &= \frac{1}{2}mv^2\\&=\frac{1}{2}m(0^2)\\&=0 \mbox{ J}\end{align}
Hence, according to the principle of conservation of energy
\begin{align}U_i+K_i=U_f+K_f\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0+\frac{1}{2}mu^2=mgh+0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}u^2=gh\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h=\frac{u^2}{2g}\end{align}
as required.
