3.2 Dynamics and Newton's laws of motion — Practise

Work through these on your own. For each question, draw a free-body diagram before you write an equation. Mark the positive direction on the diagram.

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

A resultant force of 45 N acts on an object of mass 15 kg. Find the acceleration.
A car of mass 600 kg has a driving force of 2000 N. A drag force of 800 N acts against the motion. Find the acceleration.
Two horizontal forces act on a 5.0 kg trolley: 20 N to the right and 35 N to the left. Find the resultant force and the acceleration.
A passenger of mass 60 kg stands in a lift that accelerates upwards at $2.0\,\text{m}\,\text{s}^{-2}$ . Find the normal reaction on the passenger. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A passenger of mass 70 kg stands in a lift that accelerates downwards at $1.5\,\text{m}\,\text{s}^{-2}$ . Find the normal reaction on the passenger. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A block is released on a frictionless slope at 30° to the horizontal. Find its acceleration down the slope. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A crate of mass 4.0 kg is on a slope at 25° to the horizontal. A friction force of 6.0 N acts up the slope. Find the acceleration of the crate down the slope. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A ball is thrown at $25\,\text{m}\,\text{s}^{-1}$ at 50° above the horizontal. Find the maximum height it reaches. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A stone is thrown at $18\,\text{m}\,\text{s}^{-1}$ at 35° above level ground. Find the time of flight and the range. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A skydiver of mass 90 kg falls downwards with an acceleration of $1.5\,\text{m}\,\text{s}^{-2}$ . Find the drag force on the skydiver. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A book rests on a table. A student says the weight of the book and the normal reaction from the table are a third-law pair. Explain why the student is wrong.
A ball is kicked at 22 m/s at 28 degrees above the horizontal. Say how you would set up the calculation before you find the range, and name the one quantity that is the same in both directions.

Answers and concise solutions

1. The acceleration

Answer: $a = \dfrac{F}{m} = \dfrac{45}{15} = 3.0\,\text{m}\,\text{s}^{-2}$ . The acceleration is in the direction of the resultant force.

2. The acceleration

Answer: The resultant is $2000 - 800 = 1200\,\text{N}$ , so $a = \dfrac{1200}{600} = 2.0\,\text{m}\,\text{s}^{-2}$ .

3. The resultant force and the acceleration

Answer: Taking right as positive, the resultant is $20 - 35 = -15\,\text{N}$ , so it is 15 N to the left. The acceleration is $a = \dfrac{-15}{5.0} = -3.0\,\text{m}\,\text{s}^{-2}$ , so it is $3.0\,\text{m}\,\text{s}^{-2}$ to the left.

4. The normal reaction

Answer: The resultant is upwards, so $N - mg = ma$ and $N = m(g + a) = 60 \times 11.81 = 709\,\text{N}$ to three significant figures. The weight is 589 N, so the passenger feels heavier.

5. The normal reaction

Answer: The resultant is downwards, so $mg - N = ma$ and $N = m(g - a) = 70 \times 8.31 = 582\,\text{N}$ to three significant figures. The weight is 687 N, so the passenger feels lighter.

6. The acceleration down the slope

Answer: The only force along the slope is $mg\sin 30^\circ$ , so $a = g\sin 30^\circ = 9.81 \times 0.5 = 4.9\,\text{m}\,\text{s}^{-2}$ .

7. The acceleration down the slope

Answer: The component of the weight down the slope is $4.0 \times 9.81 \times \sin 25^\circ = 16.6\,\text{N}$ . The resultant is $16.6 - 6.0 = 10.6\,\text{N}$ down the slope, so $a = \dfrac{10.6}{4.0} = 2.6\,\text{m}\,\text{s}^{-2}$ .

8. The maximum height

Answer: The vertical component is $u_y = 25\sin 50^\circ = 19.2\,\text{m}\,\text{s}^{-1}$ . At the highest point $v_y = 0$ , so $h = \dfrac{u_y^2}{2g} = \dfrac{19.2^2}{19.62} = 18.7\,\text{m}$ .

9. The time of flight and the range

Answer: The vertical component is $u_y = 18\sin 35^\circ = 10.3\,\text{m}\,\text{s}^{-1}$ , so the time of flight is $t = \dfrac{2u_y}{g} = \dfrac{20.6}{9.81} = 2.10\,\text{s}$ . The horizontal component is $u_x = 18\cos 35^\circ = 14.7\,\text{m}\,\text{s}^{-1}$ , so the range is $u_x t = 14.7 \times 2.10 = 30.9\,\text{m} \approx 31\,\text{m}$ .

10. The drag force

Answer: The resultant is $mg - D = ma$ , so $D = m(g - a) = 90(9.81 - 1.5) = 748\,\text{N}$ to three significant figures.

11. Why the two forces are not a third-law pair

The weight and the normal reaction both act on the same object, the book. A third law pair always acts on two different objects.
The third law partner of the weight of the book is the gravitational pull of the book on the Earth. The weight and the reaction balance, so the book is in equilibrium, but they are not a third-law pair.

Answer: A third-law pair acts on two different objects. The weight and the normal reaction both act on the book, so they are a balanced pair, not a third-law pair.

12. The method and the shared quantity

Resolve the initial velocity into a horizontal component u cos 28 and a vertical component u sin 28.
The horizontal velocity is constant, because there is no horizontal force. The time is the same in both directions.
Use v = u + at vertically to find the time of flight, then multiply the horizontal velocity by that time to find the range.

Answer: Resolve the velocity first. The time is the same in both directions. Use the vertical motion to find the time of flight, then multiply by the constant horizontal velocity to find the range.