2022 A/L Physics Past Paper · Question 27 (English Medium)

Answer: (3) Gravitational force: $mgR$, normal reaction force: $0$

Given information

What we know:

The sphere starts at rest on the rim of a hemispherical bowl of radius $R$. After many swings it stops at the bottom.

Only two forces are asked about: its weight (gravitational force) $mg$ and the normal reaction $N$ from the surface of the bowl.

The question has no numbers. So we keep the letters and check with sample values $m = 2\ \text{kg}$, $R = 0.5\ \text{m}$, $g = 10\ \text{m s}^{-2}$.

What the question asks

We want: the work done by gravity and the work done by the normal reaction between the start and the end.

What each letter means:

$m$ ($\text{kg}$): the mass of the sphere.

$R$ ($\text{m}$): the radius of the bowl.

$g$ ($\text{m s}^{-2}$): the acceleration due to gravity.

$N$ ($\text{N}$): the normal reaction of the bowl on the sphere.

$W_g$, $W_N$ ($\text{J}$): the work done by gravity and by the normal reaction.

Relevant theory

If the force is acting at an inclination to the displacement, then the work done is equal to the product of the component of the force in the direction of the displacement and the displacement. work done $= F\cos\theta \cdot s$ $= F.s\cos\theta$

Suppose that a body of mass m is raised from position A to a position B at a height h. The force of gravitational attraction acting on the body is $F = mg$ vertically downwards. Work has to be done against this force and hence this work is $W = F.s \Rightarrow W = mgh$ This work gets stored in the body as gravitational potential energy.

When the body is lowered from position A to a position C at a height h energy is released.

For example, push a book along a table. The table holds the book up, at right angles to the way the book moves. That push never carries the book forward and never holds it back.

The same happens here. A force does work only along the direction in which the body moves. If $\theta$ is the angle between the force and the displacement:

$$W = F\,s\cos\theta$$

In words: work is the force times the displacement times the cosine of the angle between them.

For the normal reaction $\theta = 90^\circ$, and $\cos 90^\circ = 0$. So its work is zero.

For gravity, lifting a body through a height $h$ needs the work $mgh$ against gravity. Lowering it through the same height gives that energy back. So gravity does the work $mgh$ on a body that goes down a height $h$.

$$W_g = mg\,h$$

In words: the work done by gravity is the weight times the height lowered. It does not depend on the path taken.

Required equations

$W = F\,s\cos\theta$

$W_N = N\,s\cos 90^\circ = 0$

$W_g = mg\,h = mgR$

Applying it to this question

Mass in $\text{kg}$, $g$ in $\text{m s}^{-2}$ and $R$ in $\text{m}$ give $\text{kg m s}^{-2}\,\text{m} = \text{N m} = \text{J}$. So $mgR$ is a work.

$0.5$

$h = R = 0.5$

$10$

$W_g = mgR = 2 \times 10 \times 0.5$

$0$

$W_N = N\,s\cos 90^\circ = N\,s \times 0$

Reasoning and calculation

Gravity does $mgR$ and the normal reaction does $0$. The friction that stops the sphere is not asked about, and it does not change the work of these two forces.

That is option (3).

Checks

$m * g * r$

Does it make sense?

The sphere ends at rest, lower than it began. Its potential energy has fallen by $mgR$. That energy was turned into heat by friction. So gravity did $mgR$ of work.

The bowl pushes straight out from the surface. The sphere slides along the surface. So the bowl's push does no work.

Conclusion

$W_g = mgR,\ W_N = 0$

Gravity does $W_g = mgR$ and the normal reaction does $W_N = 0$.

The answer is (3).

A force at right angles to the motion does no work. The normal reaction of a surface is like this. So is the tension of a string that swings a ball in a circle, because the string stays at right angles to the motion.

Work by gravity depends only on the height lowered, not on the path.

Exam tip