Two 10 kilogram boxes are connected by a massless string that passes over a massless frictionless pulley as shown above. 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 Ieft 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.
The Attempt at a Solution
a. T= 93.3N
b. I drew:
normal force = 100cos(60)
friction force = Fnormal x static friction coefficient
mg = 100N
c. static Ffric = 15N ... KE Ffric = 13N
Feedbacks or corrections is appreciated but I think I got everything right =D I THINK!