16.3 Weight and motion under gravity — Practise with help

Work each one first, closed book. Ask whether the question wants a mass in kilograms or a force in newtons. Take the direction of motion as positive, and write $g = 9.8$ m s $^{-2}$ unless the question says otherwise.

Warm-up: remember

Work out $25 \times 9.8$ .
Rearrange $W = mg$ to make $m$ the subject.
A resultant force of $30$ N acts on a mass of $6$ kg. Find the acceleration.

Answers

$25 \times 9.8 = 245$ .
$m = \dfrac{W}{g}$ .
$a = \dfrac{F}{m} = \dfrac{30}{6} = 5$ m s $^{-2}$ .

If question 3 was hard, revise Newton's second law in Section 16.2 before you continue.

Practise with help

1. Find the weight of a suitcase of mass $25$ kg. Take $g = 9.8$ m s $^{-2}$ .

Hint

Weight is a force, so the answer is in newtons. Use $W = mg$ .

Next step

Multiply the mass by $9.8$ , and say that the weight acts downwards.

Full working

W = mg = 25 \times 9.8 = 245 \text{ N}

acting downwards. The mass stays $25$ kg.

2. A body has a weight of $735$ N. Find its mass. Take $g = 9.8$ m s $^{-2}$ .

Hint

This time the mass is the unknown, so divide.

Next step

Use $m = \dfrac{W}{g}$ and divide $735$ by $9.8$ .

Full working

m = \frac{W}{g} = \frac{735}{9.8} = 75 \text{ kg}.

3. A bag of mass $2$ kg is taken to the Moon, where $g = 1.6$ m s $^{-2}$ . Find its mass and its weight on the Moon.

Hint

The mass does not change when the bag moves. Only the weight changes.

Next step

The mass is still $2$ kg. Use $W = mg$ with the Moon value of $g$ .

Full working

Mass $= 2$ kg.

W = mg = 2 \times 1.6 = 3.2 \text{ N}.

The bag weighs less on the Moon, but its mass is unchanged.

4. A ball of mass $0.4$ kg falls with an air resistance of $0.6$ N. Find its acceleration. Take $g = 9.8$ m s $^{-2}$ .

Hint

Take downwards as positive. The weight acts down and the air resistance acts up.

Next step

The weight is $0.4 \times 9.8 = 3.92$ N. The resultant is $3.92 - 0.6$ N. Then divide by the mass.

Full working

a = \frac{0.4 \times 9.8 - 0.6}{0.4} = \frac{3.32}{0.4} = 8.3 \text{ m s}^{-2}

downwards.

5. A skydiver of mass $75$ kg falls at terminal velocity. Find the air resistance. Take $g = 9.8$ m s $^{-2}$ .

Hint

At terminal velocity the velocity is constant.

Next step

A constant velocity means zero resultant force, so the air resistance balances the weight.

Full working

R = mg = 75 \times 9.8 = 735 \text{ N}

acting upwards, balancing the weight.

6. A person of mass $65$ kg stands in a lift that accelerates upwards at $2$ m s $^{-2}$ . Find the normal reaction from the floor. Take $g = 9.8$ m s $^{-2}$ .

Hint

The acceleration is upwards, so the reaction is larger than the weight.

Next step

Use $R = m(g + a)$ .

Full working

R = 65(9.8 + 2) = 65 \times 11.8 = 767 \text{ N}.

7. A person of mass $65$ kg stands in a lift that accelerates downwards at $2$ m s $^{-2}$ . Find the normal reaction from the floor. Take $g = 9.8$ m s $^{-2}$ .

Hint

The acceleration is downwards now, so the reaction is smaller than the weight.

Next step

Use $R = m(g - a)$ for a downward acceleration.

Full working

R = 65(9.8 - 2) = 65 \times 7.8 = 507 \text{ N}.

The person feels lighter than usual.

8. A person of mass $65$ kg stands in a lift that moves at a constant velocity. Find the normal reaction from the floor. Take $g = 9.8$ m s $^{-2}$ .

Hint

A constant velocity means no acceleration, so this case is the same as standing still.

Next step

Use $R = mg$ .

Full working

R = mg = 65 \times 9.8 = 637 \text{ N}.

The reading is the same as when the lift is at rest.

9. A stone is dropped and falls with air resistance negligible. Explain why its acceleration does not depend on its mass.

Hint

Write down the resultant force on the stone, then put it into $F = ma$ .

Next step

The only force is the weight $mg$ . Substitute that for $F$ in $F = ma$ .

Full working

The resultant force is the weight, $F = mg$ . Newton's second law gives

mg = ma,

and dividing both sides by $m$ gives $a = g$ . The mass cancels, so every freely falling body has the same acceleration.