Question

A constant force 3.5 N acts on a 7kg block, initially at rest, for 7.0 s. Find the (a) acceleration of the block, (b) speed of the block at the end of 7.0 s, (c) change in momentum of the block, and (d) impulse produced by the force

Answer 1

a) We can find the acceleration of the block from the Newton's Second Law of Motion:

$$F=ma,$$ $$a=\dfrac{F}{m}=\dfrac{3.5\ N}{7\ kg}=0.5\ \dfrac{m}{s^2}.$$

b) We can find the speed of the block at the end of 7.0 s from the kinematic equation:

$$v=v_0+at = 0\ \dfrac{m}{s}+0.5\ \dfrac{m}{s^2}\cdot 7.0\ s=3.5\ \dfrac{m}{s}.$$

c) We can find the change in momentum of the block from the definition of the momentum:

$$\Delta p=m\Delta v=7\ kg\cdot 3.5\ \dfrac{m}{s}=24.5\ \dfrac{kgm}{s}.$$

d) We can find the impulse produced by the force from the definition of the impusle:

$$J=F\Delta t=3.5\ N\cdot 7.0\ s=24.5\ N\cdot s.$$

Answer:

a) $a=0.5\ \dfrac{m}{s^2}.$

b) $v=3.5\ \dfrac{m}{s}.$

c) $\Delta p=24.5\ \dfrac{kgm}{s}.$

d) $J=24.5\ N\cdot s.$