3.2 Dynamics and Newton's laws of motion — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

A car accelerates uniformly from $12\,\text{m}\,\text{s}^{-1}$ to $30\,\text{m}\,\text{s}^{-1}$ in 6.0 s. Find its acceleration.
A stone falls from rest for 1.5 s. Find its speed. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A force of 40 N acts at 25° to the horizontal. Find its horizontal and vertical components.

Answers

$a = \dfrac{30 - 12}{6.0} = 3.0\,\text{m}\,\text{s}^{-2}$ .
$v = gt = 9.81 \times 1.5 = 14.7\,\text{m}\,\text{s}^{-1}$ .
$40\cos 25^\circ = 36.3\,\text{N}$ and $40\sin 25^\circ = 16.9\,\text{N}$ .

If question 1 goes wrong, use $a = \dfrac{v - u}{t}$ . If question 2 goes wrong, the stone starts from rest and the acceleration is $g$ . If question 3 goes wrong, the angle is measured from the horizontal, so that component uses the cosine.

Practise with help

1. A resultant force of 36 N acts on an object of mass 12 kg. Find the acceleration.

Hint

The force given is already the resultant, so it can be used directly in $F = ma$ .

Next step

Rearrange $F = ma$ to make the acceleration the subject.

Full working

a = \frac{F}{m} = \frac{36}{12} = 3.0\,\text{m}\,\text{s}^{-2}

The acceleration is in the same direction as the resultant force.

2. A car of mass 1500 kg has a driving force of 4000 N. A drag and friction force of 2500 N acts against the motion. Find the acceleration.

Hint

Take the direction of motion as positive. The 4000 N drives forwards and the 2500 N acts backwards.

Next step

Find the resultant force first, then divide by the mass.

Full working

\begin{aligned} F &= 4000 - 2500 = 1500\,\text{N} \\ a &= \frac{1500}{1500} = 1.0\,\text{m}\,\text{s}^{-2} \end{aligned}

The acceleration is $1.0\,\text{m}\,\text{s}^{-2}$ in the direction of motion.

3. A crate of mass 20 kg is released on a frictionless slope at 15° to the horizontal. Find its acceleration down the slope and the normal reaction on it. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

Resolve the weight into a component along the slope and a component perpendicular to it. Only the along-slope component accelerates the crate.

Next step

The along-slope component is $mg\sin 15^\circ$ , so $a = g\sin 15^\circ$ . The perpendicular component is balanced by the normal reaction.

Full working

\begin{aligned} a &= g\sin 15^\circ = 9.81 \times 0.259 = 2.5\,\text{m}\,\text{s}^{-2} \\ N &= mg\cos 15^\circ = 20 \times 9.81 \times 0.966 = 190\,\text{N} \end{aligned}

The acceleration does not depend on the mass: doubling the mass doubles both the force and the inertia. The normal reaction is less than the weight, because the slope is tilted.

4. A passenger of mass 80 kg stands in a lift that accelerates downwards at $0.80\,\text{m}\,\text{s}^{-2}$ . Find the normal reaction on the passenger.

Hint

The acceleration is downwards, so the weight is larger than the normal reaction. The resultant is $mg - N$ downwards.

Next step

Write Newton's second law downwards: $mg - N = ma$ , then rearrange for $N$ .

Full working

\begin{aligned} mg - N &= ma \\ N &= m(g - a) \\ &= 80(9.81 - 0.80) \\ &= 721\,\text{N} \end{aligned}

The weight is $785\,\text{N}$ and the normal reaction is $721\,\text{N}$ , so the passenger feels lighter while the lift accelerates downwards.

5. A ball is thrown at $15\,\text{m}\,\text{s}^{-1}$ at 40° above the horizontal. Find the maximum height, the time of flight and the range. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

Resolve the launch velocity into a horizontal component and a vertical component. The horizontal component stays constant.

Next step

At the highest point the vertical velocity is zero, so use $v_y^2 = u_y^2 - 2gh$ . The time of flight is twice the time to the highest point.

Full working

\begin{aligned} u_x &= 15\cos 40^\circ = 11.5\,\text{m}\,\text{s}^{-1} \\ u_y &= 15\sin 40^\circ = 9.64\,\text{m}\,\text{s}^{-1} \\ h_{\max} &= \frac{u_y^2}{2g} = \frac{9.64^2}{19.62} = 4.7\,\text{m} \\ t_{\text{top}} &= \frac{u_y}{g} = \frac{9.64}{9.81} = 0.983\,\text{s} \\ t_{\text{flight}} &= 2 \times 0.983 = 1.97\,\text{s} \\ \text{range} &= u_x t_{\text{flight}} = 11.5 \times 1.97 = 23\,\text{m} \end{aligned}

The range is $23\,\text{m}$ to two significant figures.

6. A raindrop falls from a cloud and eventually moves at a constant speed. Explain how the forces on the drop lead to this terminal velocity, and say what the resultant force is at terminal velocity.

Hint

Name the two forces on the drop. One is constant and the other grows with speed.

Next step

Describe what happens to the resultant force, and so to the acceleration, as the speed rises.

Full working

Two forces act on the drop: its weight downwards, which is constant, and the drag upwards, which grows as the drop speeds up.

At first the drop is slow, so the drag is small. The resultant is close to the weight, and the acceleration is close to $g$ . As the drop speeds up, the drag grows. The resultant force falls, so the acceleration falls. When the drag becomes equal to the weight, the resultant is zero and the drop stops accelerating. It then falls at a constant terminal velocity.

At terminal velocity the resultant force is zero, so the net force is zero and the drop moves at constant speed.