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

Work through these on your own. Choose the direction of motion as positive. Write the resultant force along that direction, then use $F = ma$ . Check the units: mass in kilograms, acceleration in metres per second squared, force in newtons.

Both 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 $20$ N acts on a mass of $5$ kg. Find the acceleration.
A car of mass $1200$ kg brakes with a total braking force of $3600$ N. The only horizontal force is the braking force. Find the acceleration, taking the direction of motion as positive.
A crate of mass $45$ kg is pushed along a floor by a resultant force of $150$ N. Find the acceleration.
A mass of $8$ kg accelerates at $2.5$ m s $^{-2}$ . Find the resultant force.
A resultant force of $240$ N gives a body an acceleration of $4$ m s $^{-2}$ . Find the mass of the body.
Forces of $50$ N and $30$ N act in opposite directions on a body of mass $8$ kg. Find the acceleration.
A car of mass $1500$ kg has a driving force of $4000$ N and a total resistance of $1000$ N. Find the acceleration.
A body of mass $4$ kg accelerates from rest to $20$ m s $^{-1}$ in $5$ s. Find the resultant force on it.
A car of mass $900$ kg must decelerate at $2$ m s $^{-2}$ . Find the braking force needed if the only horizontal force is the braking force.
A resultant force of $(6, 8)$ N acts on a mass of $2$ kg. Find the acceleration as a column vector, and its size.
A body of mass $6$ kg accelerates at $1.5$ m s $^{-2}$ against a resistance of $5$ N. Find the driving force.
Forces of $30$ N and $40$ N act at right angles on a mass of $10$ kg. Find the size of the resultant force and the acceleration.

Answers and concise solutions

1. The acceleration

Answer: Use $F = ma$ with $F = 20$ and $m = 5$ : $a = \dfrac{F}{m} = \dfrac{20}{5} = 4$ m s $^{-2}$ . The acceleration is in the direction of the resultant force.

2. The acceleration

Answer: The braking force acts backwards, so $F = -3600$ N and $a = \dfrac{-3600}{1200} = -3$ m s $^{-2}$ . The negative sign means the car decelerates at $3$ m s $^{-2}$ .

3. The acceleration

Answer: $a = \dfrac{F}{m} = \dfrac{150}{45} = \dfrac{10}{3} \approx 3.33$ m s $^{-2}$ . Leave the answer as a fraction, or round it at the end.

4. The resultant force

Answer: Rearrange $F = ma$ for the force: $F = 8 \times 2.5 = 20$ N.

5. The mass

Answer: From $F = ma$ , $m = \dfrac{F}{a} = \dfrac{240}{4} = 60$ kg.

6. The acceleration

Answer: The resultant force is $50 - 30 = 20$ N in the direction of the $50$ N force. Then $a = \dfrac{20}{8} = 2.5$ m s $^{-2}$ .

7. The acceleration

Answer: The resultant force is $4000 - 1000 = 3000$ N. Then $a = \dfrac{3000}{1500} = 2$ m s $^{-2}$ .

8. The resultant force

Answer: First find the acceleration: $a = \dfrac{20 - 0}{5} = 4$ m s $^{-2}$ . Then $F = ma = 4 \times 4 = 16$ N.

9. The braking force

Answer: A deceleration of $2$ m s $^{-2}$ needs a resultant force of $F = ma = 900 \times 2 = 1800$ N acting backwards. The braking force is $1800$ N.

10. The acceleration and its size

Answer: Divide each component of the force by the mass: $\mathbf{a} = \dfrac{1}{2}(6, 8) = (3, 4)$ m s $^{-2}$ . Its size is $\sqrt{3^2 + 4^2} = 5$ m s $^{-2}$ .

11. The driving force

Answer: The resultant force must be $F = ma = 6 \times 1.5 = 9$ N. The driving force also has to overcome the $5$ N resistance, so it is $9 + 5 = 14$ N.

12. The resultant and the acceleration

Answer: The resultant force is $\sqrt{30^2 + 40^2} = 50$ N. The acceleration is $a = \dfrac{50}{10} = 5$ m s $^{-2}$ , in the direction of the resultant.