Question

A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees and the coefficient of dynamic friction is 0.72, what is the speed with which the box is moving with, assuming it takes 4seconds to reach the top of the incline?

Answer 1

Given :

Mass of box , m = 250 kg.

Force applied , F = 285 N.

The value of the incline angle is 30°.

the coefficient of dynamic friction is $\mu=0.72$ .

To Find :

The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.

Solution :

Net force applied in box is :

$F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N$

Acceleration , $a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2$.

By equation of motion :

$v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s$

Therefore, the speed of box is 12.04 m/s.

Hence, this is the required solution.