Question

An 30 kg object from rest was pulled horizontally by a pulling force 100 N applied along the direction of displacement line. The coefficient of kinetic friction between the surface and the object is 0.20 and the object was displaced at a distance of 40 meters. How much power consumed during the travel?

Answer 1

Let's first find the net force:

$$F_{net}=F_{pull}-F_{k.fr.},$$ $$F_{net}=F_{pull}-\mu_kmg,$$ $$F_{net}=100\ N-0.2\times30\ kg\times9.8\ N = 41.2\ N.$$

Then, we can find the net work:

$$W_{net}=F_{net}d=41.2\ N\times40\ m=1648\ J.$$

Let's find the velocity of the object at the end of the pull (when it was displaced at a distance of 40 meters). Applying the work-kinetic energy theorem, we get:

$$W_{net}=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{\dfrac{2W_{net}}{m}}=\sqrt{\dfrac{2\times1648\ J}{30\ kg}}=10.5\ \dfrac{m}{s}.$$

Then, we can find the acceleration of the object:

$$a=\dfrac{v^2}{2d}=\dfrac{(10.5\ \dfrac{m}{s})^2}{2\times40\ m}=1.38\ \dfrac{m}{s^2}.$$

The time taken to travel the distance of 40 meters can be found as follows:

$$t=\dfrac{v}{a}=\dfrac{10.5\ \dfrac{m}{s}}{1.38\ \dfrac{m}{s^2}}=7.6\ s.$$

Finally, we can find the power consumed during the travel:

$$P=\dfrac{W_{net}}{t}=\dfrac{1648\ J}{7.6\ s}=217\ W.$$