16.2 Newton's second law and F = ma — Practise with help

Work each one first, closed book. Choose the direction of motion as positive. Write down the resultant force along that direction, then use $F = ma$ .

Warm-up: remember

Work out $30 - 12$ and $12 - 30$ .
Solve $5a = 20$ .
A car speeds up from $10$ m s $^{-1}$ to $22$ m s $^{-1}$ in $4$ s. Find its acceleration.

Answers

$30 - 12 = 18$ and $12 - 30 = -18$ .
$a = 4$ .
$a = \dfrac{22 - 10}{4} = 3$ m s $^{-2}$ .

If question 3 was hard, revise the constant-acceleration formulae in Section 15.4 before you continue.

Practise with help

1. Forces of $30$ N and $12$ N act in opposite directions on a mass of $8$ kg. Find the acceleration.

Hint

The resultant is the difference between the two forces, because they oppose each other.

Next step

The resultant is $30 - 12 = 18$ N in the direction of the $30$ N force. Now use $a = \dfrac{F}{m}$ .

Full working

a = \frac{30 - 12}{8} = \frac{18}{8} = 2.25 \text{ m s}^{-2}

in the direction of the $30$ N force.

2. A crate of mass $20$ kg is pushed along a horizontal floor by a force of $90$ N. A resistance of $10$ N opposes the motion. Find the acceleration.

Hint

The resistance is a force in the negative direction.

Next step

The resultant force is $90 - 10 = 80$ N. Then divide by the mass.

Full working

a = \frac{90 - 10}{20} = \frac{80}{20} = 4 \text{ m s}^{-2}.

3. A mass of $7$ kg accelerates at $3$ m s $^{-2}$ . Find the resultant force.

Hint

This time the force is the unknown in $F = ma$ .

Next step

Multiply the mass by the acceleration.

Full working

F = ma = 7 \times 3 = 21 \text{ N}.

4. A resultant force of $144$ N gives a body an acceleration of $6$ m s $^{-2}$ . Find the mass of the body.

Hint

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

Next step

The mass is the force divided by the acceleration.

Full working

m = \frac{F}{a} = \frac{144}{6} = 24 \text{ kg}.

5. A car of mass $1500$ kg has a driving force of $4000$ N and a total resistance of $1600$ N. Find the acceleration.

Hint

Find the resultant force along the road first.

Next step

The resultant is $4000 - 1600 = 2400$ N. Then use $a = \dfrac{F}{m}$ .

Full working

a = \frac{4000 - 1600}{1500} = \frac{2400}{1500} = 1.6 \text{ m s}^{-2}.

6. A car of mass $800$ kg brakes with a force of $2400$ N. Find the deceleration.

Hint

The braking force is the only horizontal force, and it points backwards.

Next step

Divide the braking force by the mass. The minus sign in the algebra stands for a deceleration.

Full working

Taking the direction of motion as positive,

a = \frac{-2400}{800} = -3 \text{ m s}^{-2}.

The car decelerates at $3$ m s $^{-2}$ .

7. A trolley of mass $5$ kg accelerates from rest to $15$ m s $^{-1}$ in $3$ s. Find the acceleration and the resultant force.

Hint

This is two steps: first find the acceleration from the motion, then use $F = ma$ .

Next step

The acceleration is $\dfrac{15 - 0}{3}$ . Multiply the answer by the mass.

Full working

a = \frac{15 - 0}{3} = 5 \text{ m s}^{-2}, \qquad F = ma = 5 \times 5 = 25 \text{ N}.

8. A body of mass $12$ kg accelerates at $0.5$ m s $^{-2}$ against a resistance of $8$ N. Find the driving force.

Hint

First find the resultant force the acceleration needs. The driving force is larger than the resultant when a resistance is present.

Next step

The resultant is $12 \times 0.5 = 6$ N. Add the resistance to find the driving force.

Full working

F_{\text{resultant}} = ma = 12 \times 0.5 = 6 \text{ N},

so the driving force is $6 + 8 = 14$ N.

9. Forces of $6$ N and $8$ N act at right angles on a mass of $4$ kg. Find the acceleration.

Hint

The two forces are perpendicular, so find their resultant with Pythagoras before you use $F = ma$ .

Next step

The resultant force is $\sqrt{6^2 + 8^2} = 10$ N. Divide by the mass.

Full working

F = \sqrt{6^2 + 8^2} = 10 \text{ N}, \qquad a = \frac{10}{4} = 2.5 \text{ m s}^{-2},

in the direction of the resultant force.