Question

A 2.0 kg mass is released from rest at the top of a plane at 20 degrees above the horizontal. The coefficient of kinetic friction between the mass and the plane is 0.20. What will be the speed of the mass after sliding 4.0 m along the plane?

Answer 1

Answer:

3.97 m/s

Explanation:

given,

mass of the object,m= 2 Kg

angle of inclination,θ = 20°

coefficient of friction,μ= 0.2

distance,L = 4 m

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

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

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

 $v= \sqrt{2(9.8\times sin 20^0 - 0.15\times 9.8 \times cos 20^0)4}$

     v = 3.97 m/s

hence, the final speed of the box is equal to v = 3.97 m/s