Question

A 50kg object is moving at 18.2 m/s when a 200 N force is applied in the opposite direction of it’s motion. If the object slows down to 12.6 m/s due to the force, how long was the force applied?

Answer 1

According to the work-energy theorem, the change in the kinetic energy is equal to the work of the force applied to the object:

$$\dfrac{mv_2^2}{2} - \dfrac{mv_1^2}{2} = W$$

where $m = 50kg$ is the mass of the object, $v_1 = 18.2 m/s, v_2 = 12.6m/s$ are its initial and final speeds respectively, $W = Fs$ is the work done by the force $F = 200N$ acting along the distance $s$. Expressing $s$, find:

$$\dfrac{mv_2^2}{2} - \dfrac{mv_1^2}{2} = Fs\\ s = \dfrac{m}{2F}(v_2^2 - v_1^2)$$

On the other hand, the distance traveled under a constant force (constant acceleration) is given as follows:

$$s = v_1t - \dfrac{at^2}{2}$$

where $a = \dfrac{F}{m}$ is the acceleration of the body (from the second Newton's law) and $t$ is the time of the motion. Combining all together, obtain:

$$\dfrac{m}{2F}(v_2^2 - v_1^2) = v_1t - \dfrac{Ft^2}{2m}\\ \dfrac{Ft^2}{2m} - v_1t + \dfrac{m}{2F}(v_2^2 - v_1^2) = 0$$

Substituting the numbers and solving the quadratic equation for $t$, obtain:

$$2t^2 - 18.2t - 21.56 = 0\\ t_1 = -1.06s, t_2 = 10.2s$$

Taking the positive value (since time can not be negative), obtain the final answer.

Answer. 10.2s.