Question

A box is placed at the top of an incline at an angle 22.9 degrees to the horizontal. When released the box slides 12.1 meters down the incline. If the kinetic coefficient of friction between the box and the incline is 0.15, what is the final speed of the box?

Answer 1

Answer:

v = 7.71 m/s

Explanation:

given,

angle of inclination = 22.9°

distance moved = 12.1 m

coefficient of friction = 0.15

final speed = ?

using work energy theorem

$W = \dfrac{1}{2}mv^2$..........(1)

now,

v   is the speed of the box.

W  is the work done on the box by a net external force.

Work done on the force

$W = F_{net}L$

$F_{net} = m g sin \theta - \mu N$

$F_{net} = m g sin \theta - \mu (mg cos \theta)$

$W = (mg sin\theta - \mu mg cos \theta)L$......(2)

now, equating both equation 1 and 2

$\dfrac{1}{2}mv^2 = (mg sin\theta - \mu mg cos \theta)L$

 $v= \sqrt{2 (g sin\theta - \mu g cos \theta)L}$

 $v= \sqrt{2(9.8\times sin 22.9^0 - 0.15\times 9.8 \times cos 22.9^0)12.1}$

     v = 7.71 m/s

hence, the final speed of the box is equal to v = 7.71 m/s