Two forces, of magnitudes F1 = 95.0 N and F2 = 40.0 N , act in opposite directions on a block, which sits atop a frictionless surface

Question

Initially, the center of the block is at position  = -6.00  . At some later time, the block has moved to the right, and its center is at a new position,  = 6.00  .

Find the work done on the block by the force of magnitude  F1= 95.0  as the block moves from  xi= -6.00  to  xf=  6.00 

Find the work  done by the force of magnitude F2  = 40.0  as the block moves from  xi= -6.00  to  xf = 6.00 

What is the net work  done on the block by the two forces?

Determine the change in the kinetic energy of the block as it moves from xi = -6.00  to  xf= 6.00  

Answer 1

Work done by F1 = 95N acting alone = 95N x (6.00m - -6.00m) m =95N x 12 m = 1140 joules.

Work done by F2 = -40N x 12 m = -40N - (6.00m - -6m) = -40N x 12m) = -480 joules.

Net work done by the forces. 1140N + -480J = 660 j

If the block was initially at rest, the kinetic energy is zero, the net work must be equal to the final kinetic energy change (660 j).