Question

A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.42 m/s. The pulling force is 96N parallel to the incline, which makes an angle of 20.4 with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.02 m.

A) how much work is done by the gravitational force on the crate?

B) determine the increase in internal energy of the crate-incline system owing to friction?

C) how much work is done by the 96-N force on the crate?

D) what is the change in kinetic energy of the crate?

E) what is the speed of the crate after being pulled 5.02 m?

Thanks

Answer 1

a)

Since Parallel component "mgSintheta" and 'd' are in opposite direction , angle between them is 180

Work done by the gravitational force is given as :

W = mgh = mg Sintheta x d Cos180 = 9.2 x 9.8 Sin20.4 (5.02) Cos180 = - 157.8 J

b)

using equilibrium of forces along perpendicular direction to incline

F_n = mg Cos20.4 = 9.2 x 9.8 Cos20.4 = 84.5 N

Frictional force is given as :

f_k = u_k F_n = 0.4 x 84.5 = 33.8 N

Work done by frictional force is :

W_f = f_k d Cos180 = 33.8 x 5.02 Cos180 = - 169.7 J       (Since f_k is opposite to 'd' )

C)

Work done by 96 N force :

W = 96 x d Cos0 = 96 x 5.02 = 481.9 J            (since 96 N acts in the same direction as 'd')

D)

Net force on the crate = F_net = F - f_k - mg Sintheta = 96 - 33.8 - 9.2 x 9.8 Sin20.4 = 30.77 N

acceleration = a = F_net /m = 30.77 / 9.2 = 3.345 m/s²

using the equation

V_f² = V_i² + 2 a d

multiplying Both side by 0.5m

(0.5m)V_f² = (0.5m)V_i² + 2 a d(0.5m)

change in KE = mad = 30.77 x 5.02 = 154.5 J