16.3 Weight and motion under gravity — Practise

Work through these on your own. Decide first whether the answer is a mass or a force, and use the right unit. Unless a question says otherwise, take $g = 9.8$ m s $^{-2}$ .

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Find the weight of a crate of mass $45$ kg. Take $g = 9.8$ m s $^{-2}$ .
Find the weight of a person of mass $70$ kg. Take $g = 9.8$ m s $^{-2}$ .
A body has a weight of $588$ N. Find its mass. Take $g = 9.8$ m s $^{-2}$ .
Find the weight of a book of mass $2.5$ kg. Take $g = 9.8$ m s $^{-2}$ .
A ball of mass $0.5$ kg is released and falls. Air resistance is negligible. Find the resultant force on the ball and its acceleration. Take $g = 9.8$ m s $^{-2}$ .
A skydiver of mass $60$ kg falls with an air resistance of $200$ N. Find the acceleration. Take $g = 9.8$ m s $^{-2}$ .
A skydiver of mass $60$ kg falls at terminal velocity. Find the air resistance. Take $g = 9.8$ m s $^{-2}$ .
A person of mass $70$ kg stands on the floor of a lift. The lift accelerates upwards at $1.5$ m s $^{-2}$ . Find the normal reaction from the floor. Take $g = 9.8$ m s $^{-2}$ .
A person of mass $70$ kg stands on the floor of a lift. The lift accelerates downwards at $1.5$ m s $^{-2}$ . Find the normal reaction from the floor. Take $g = 9.8$ m s $^{-2}$ .
A person of mass $70$ kg stands on the floor of a lift moving at a constant velocity. Find the normal reaction from the floor. Take $g = 9.8$ m s $^{-2}$ .
An astronaut of mass $45$ kg is on the Moon, where $g = 1.6$ m s $^{-2}$ . Find the astronaut's weight on the Moon.
A ball of mass $0.2$ kg falls with an air resistance of $0.5$ N. Find its acceleration. Take $g = 9.8$ m s $^{-2}$ .

Answers and concise solutions

1. The weight

Answer: Use $W = mg$ : $W = 45 \times 9.8 = 441$ N, acting downwards. The mass stays $45$ kg; only the weight is a force.

2. The weight

Answer: $W = mg = 70 \times 9.8 = 686$ N acting downwards.

3. The mass

Answer: Rearrange $W = mg$ : $m = \dfrac{W}{g} = \dfrac{588}{9.8} = 60$ kg. Mass is measured in kilograms, not newtons.

4. The weight

Answer: $W = mg = 2.5 \times 9.8 = 24.5$ N, acting downwards.

5. The resultant force and the acceleration

Answer: The only force is the weight, so $F = mg = 0.5 \times 9.8 = 4.9$ N downwards. Then $a = \dfrac{F}{m} = \dfrac{4.9}{0.5} = 9.8$ m s $^{-2}$ . A freely falling body accelerates at $g$ .

6. The acceleration

Answer: The weight is $60 \times 9.8 = 588$ N downwards and the air resistance is $200$ N upwards. The resultant is $588 - 200 = 388$ N downwards, so $a = \dfrac{388}{60} \approx 6.47$ m s $^{-2}$ downwards.

7. The air resistance

Answer: At terminal velocity the acceleration is zero, so the resultant force is zero. The air resistance balances the weight: $60 \times 9.8 = 588$ N upwards.

8. The normal reaction

Answer: Taking upwards as positive, $R - mg = ma$ , so $R = m(g + a) = 70(9.8 + 1.5) = 791$ N. The reaction is larger than the weight, so the person feels heavier.

9. The normal reaction

Answer: Taking downwards as positive, $mg - R = ma$ , so $R = m(g - a) = 70(9.8 - 1.5) = 581$ N. The reaction is smaller than the weight, so the person feels lighter.

10. The normal reaction

Answer: A constant velocity means the acceleration is zero, so the resultant force is zero. The normal reaction equals the weight: $70 \times 9.8 = 686$ N. The reading is the same as when the lift is at rest.

11. The weight on the Moon

Answer: The mass does not change, and it is still $45$ kg. The weight is $W = mg = 45 \times 1.6 = 72$ N. The smaller value of $g$ gives a smaller weight.

12. The acceleration

Answer: The weight is $0.2 \times 9.8 = 1.96$ N downwards. The resultant is $1.96 - 0.5 = 1.46$ N downwards, so $a = \dfrac{1.46}{0.2} = 7.3$ m s $^{-2}$ . The air resistance reduces the acceleration below $g$ .