Home / CIE AS & A Level Physics : 5.2 Gravitational potential energy and kinetic energy – Exam style question – Paper 1
The diagram shows a block \(P\) of mass \(M\) connected by a string over a frictionless pulley to a block \(Q\) of mass \(m\).
Block \(P\) moves up the slope with a constant velocity \(v\). The slope is at an angle \(\theta\) to the horizontal.
The acceleration due to free fall is \(g\).
The resistive forces on the blocks are negligible.
Which expressions give the energy transferred per unit time to block \(P\)?
1. \(Mgv\)
2. \(Mgv\sin\theta\)
3. \((M+m)gv\)
Correct Answer: \( \boxed{\mathrm{A}} \)
The power delivered to block \(P\) by the tension is \(Tv\). Since the block moves with constant velocity, \(T=Mg\sin\theta\).
Hence, the rate of increase of gravitational potential energy of \(P\) is \(Mg(v\sin\theta)=Mgv\sin\theta\).
Also, the power supplied by the hanging block is \(mgv\), which equals the power transmitted through the string. Thus expressions 1 and 2 are valid in the context of energy transfer.
Therefore, the correct answer is (A).
An object travels between two points. The change in gravitational potential energy \( \Delta E_p \) of the object is given by
\( \Delta E_p = mg\Delta h \)
where \(m\) is the mass of the object, \(\Delta h\) is its change in height and \(g\) is the acceleration due to free fall.
What is a necessary condition in order for the above equation to be valid?
Correct Answer: \( \boxed{\mathrm{B}} \)
The equation \( \Delta E_p=mg\Delta h \) assumes that the gravitational field strength \(g\) is constant.
This is true only in a uniform gravitational field.
Therefore, the correct answer is (B).
An electric motor operating a lift has an output power of \(20\,\mathrm{kW}\).
The lift and passengers have a combined mass of \(1500\,\mathrm{kg}\). The motor raises the lift at constant speed through a distance of \(20\,\mathrm{m}\).
How long does it take?
Correct Answer: \( \boxed{\mathrm{B}} \)
At constant speed, the work done equals the gain in gravitational potential energy.
\( W=mgh=1500\times9.8\times20=2.94\times10^{5}\,\mathrm{J} \)
Using
\( P=\dfrac{W}{t} \)
\( t=\dfrac{W}{P}=\dfrac{2.94\times10^{5}}{2.0\times10^{4}}=14.7\,\mathrm{s}\approx15\,\mathrm{s} \)
Therefore, the correct answer is (B).