Answer to Question #142445 in Physics for Ranxe

Question

An axe that weighs 8.0 kg is used to strikes the log with a force of 400 N that comes to rest in 0.04 seconds. What is the magnitude of the impulse? How fast is the axe approach the log?

Answer 1

By definition, the impulse is:

$$J = F\Delta t$$

where $F = 400N$ is the force and $\Delta t = 0.04s$ is the time of interaction. Thus:

$$J = 400N\times 0.04s = 16\space kg\cdot m/s$$

According to the second Newton's law, the acceleration of the axe is:

$$a = \dfrac{F}{m} = \dfrac{400N}{8kg} = 50m/s^2$$

where $m = 8kg$ is the mass of the axe.

On the other hand, by definition, the acceleration is:

$$a = \left| \dfrac{v_f - v_i}{\Delta t} \right|$$

where $v_f$ is the final speed of the axe and $v_i$ is the initial speed of the axe. Since the axe eventually comes to rest, its final speed is $v_f = 0$. Then the initial speed is:

$$v_i = a\Delta t = 50m/s^2\times 0.04s = 2m/s$$

Answer. $J = 16 \space kg\cdot m/s$, $v_i = 2m/s$.