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. Draw and label all 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 (when it is at rest) The string is then cut and the box slides down the 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 box when it reaches the bottom of the plane. KE= mv2/2 NonConservative Work = deltaKE + delta PE Work = Force * distance a. I said 100N, but i am not completely sure. I only said it because i thought the tension was constant throughout the whole string, and since T = mg in the hanging block because it was at rest, i said T = 100 b. i have normal force = mgcos60 and mgsin60 pointing down the slope with frictional force c. I said 13N because it is at rest and mgsin60+force of friction should equal T, but using the coeffictients of friction and the normal force i'm getting 15N ?????????? d. I said it should be 26J because of 13N times 2m, but i remembered that the frictional force would be kinetic, so FN*coefficient of kinetic = 7.5N, so the thermal energy would be 15J ???????? e. I said 148J using -26J as the WNC but i got 159J using the other frictional force as the nonconservative force.?????/??? I know how to get the answers I'm just confused as to which ones are right. Any help or insight on this would be great. Thanks.