Question

Two 10-kilogram boxes are connected by a mass less string that passes over a mass less friction less pulley as shown. The boxes remain at rest, with the one on the right hanging vertically and the one on the left 2.0 meters from the bottom of an inclined plane that makes an angle of 60° with the horizontal. The coefficients of kinetic friction and static friction between the left-hand box and the plane are 0.15 and 0.30, respectively. You may 

use g = 10 m/s² , sin 60° = 0.87, and cos 60° = 0.50.

 

a. What is the tension T in the string?

b. On the diagram below, draw and label all the forces acting on the box that is on the plane.

c. Determine the magnitude of the frictional force acting on the box on the plane. 

The string is then cut and the left-hand box slides down the inclined plane. 

d. Determine the amount of mechanical energy that is converted into thermal energy during the slide to the bottom. 

e. Determine the kinetic energy of the left-hand box when it reaches the bottom of the plane.

Answer 1

(a) The tension in the string can be found easily by isolating the 10kg mass. Only two forces act on this mass, the Tension upwards and the weight down (mg) …. Since the systems is at rest, T = mg = 100 N]

(c) Apply F_net = 0 along the plane. 

T – f_s – mg sin θ = 0 

(100 N) – f_s – (10)(10)(sin60) s

f_s = 13N

(d) Loss of mechanical energy = Work done by friction while sliding

First find kinetic friction force Perpendicular to plane 

F_net = 0

F_k = µ_k mgcosθ  F_n = mgcosθ

W_fk = f_kd = µkmg cos θ (d) = (0.15)(10)(10)(cos(60)) = 15J converted to thermal energy

(e) Using work-energy theorem … The U at the start – loss of energy from friction = K left over

U - W_fk = K

mgh - W_fk = K

mg(d sin60) – 15 = K

(10)(10)(2) sin60 – 15 = K 

K = 158J