Question

A 35 kg trunk is pushed 5.0 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic friction between the trunk and the incline is 0.16. Calculate the work done by the applied horizontal force. Calculate the work done by the weight of the trunk. How much energy was dissipated by the frictional force acting on the trunk? From now suppose the 50 kg trunk is pushed 5.4 m at constant speed up a 30° incline by a force along the plane (not as in the figure). The coefficient of kinetic friction between the trunk and the incline is 0.23. Calculate the work done by the applied force. Calculate the work done by the weight of the trunk. How much energy was dissipated by the frictional force acting on the trunk?

Answer 1

using F = m g ( sin30 + u cos30 ) / cos30 - u sin30

F = 35 X 9.8 X ( sin30 + 0.16 cos30 ) / cos30 - 0.16 sin30

F = 278.651 N

W_F = F cos30 d

= 278.65 X cos30 X 5

W_F = 1206.5898 J

--- the work done by the weight of the trunk

W = m g d sin30

W = 50 X -9.8 X 5 X sin30

W = - 1225 J

then now energy was dissipated by the frictional force acting on the trunk

W_f = u X d X ( m g cos30 + F sin30 )

W_f = 0.23 X 5 X ( 50 X 9.8 X cos30 + 278.651 Xsin30 )

W_f = 0.23 X 5 X 563.677

W_f = 648.2296 J