Sorry, I can't draw a force diagram of the box.
The sum of the forces in the y-direction is zero:
N - 20 + 30*sin(37) = 0
Solve for N, the normal force
B) The frictional force between the block and the table is defined as µk*N = 0.25*N
C) The work done by the frictional force is 0.25*N*5
D) The work done by the pushing force is 30*cos(37); The net work is 30*5*cos(37) - 0.25*N*5
A block of weight 20 N is pushed with a force of 30 N through a horizontal distance of 5 m using a stick which is at an angle of 37° from the horizontal as shown. The coefficient of kinetic friction µk between the table and the block is 0.25. For your own reference, draw a force diagram of the block.
A) What is the normal force on the block?
B) What is the frictional force between the block and the table?
C) What is the work done by the frictional force?
D) What is the net work done on the block?
1 answer
1 answer