Question

A 50 kg person launches themselves at a 65 kg box at a speed of 5.8 m/s. The force of friction between the box and the ground is 382 N. How far will the box move?

Answer 1

First find the total mass m of the person and the box

$$m={{m}_{p}}+{{m}_{b}}=50\text{ }kg+65\text{ }kg=115\,kg$$

Then we use the Newton's second law of motion to find the deceleration a due to the force F of friction between the box and the ground

$$a=-\frac{F}{m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$

The minus sign indicates that they decelerate.

If we know the initial, v_i, and final, v_f,  speeds of the person and the box, as well as their deceleration, then to find the distance s, that they travels to stop, it is best to use the equation

$$s=\frac{v_{f}^{2}-v_{i}^{2}}{2a}$$

Since v_f=0 m/s we get

$$s=\frac{-v_{i}^{2}}{2a}$$

Now use equation (1)

$$s=\frac{mv_{i}^{2}}{2F}$$

Substituting the known values, m=115 kg, v_i=5.8 m/s, F=382 N, we get

$$s=\frac{115\,kg\cdot {{5.8}^{2}}\text{ }{{m}^{2}}/{{s}^{2}}}{2\cdot 382\text{ }N}\approx 5.1\,m$$