Question

A 10.8 kg object is moving along the + x axis with a speed of 10.0 m/s . A constant force is applied to this object which gradually brings the object to rest . The object traveled 13.0 meters during this time . To the nearest 0.1 N , what was the magnitude of the force ? Do not include units in your answer .

Answer 1

The acceleration provided by the force is:

$$a= \dfrac{v_f-v_i}{t} = -\dfrac{v_i}{t}$$

where $v_f = 0m/s$ is the final speed of the object, $v_i = 10m/s$ is the initial speed, $t$ is the time the object comes to rest.

The distance travelled in this time is given as follows:

$$d = v_i t + \dfrac{at^2}{2}\\ d = v_i t - \dfrac{v_it}{2} = \dfrac{v_it}{2}$$

Since $d = 13m$, the time will be:

$$t = \dfrac{2d}{v_i}$$

Then the magnitude of the acceleration will be:

$$a = \dfrac{v_i}{t} = \dfrac{v_i^2}{2d}$$

Then the force will be (according to the second Newton's law):

$$F = ma$$

where $m = 10.8kg$ is the mass of the object. Thus, obtain:

$$F = \dfrac{mv_i^2}{2d}\\ F = \dfrac{10.8\cdot 10^2}{2\cdot 13}\approx 41.5N$$

Answer. 41.5 N.