Question

Two forces, of magnitudes F1 = 95.0 N and F2 = 45.0 N , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. (Figure 1) Initially, the center of the block is at position xi = -1.00 cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 6.00 cm .

Part A

Find the work W1 done on the block by the force of magnitude F1 = 95.0 N as the block moves from xi = -1.00 cm to xf = 6.00 cm .

Part B

Find the work W2 done by the force of magnitude F2 = 45.0 N as the block moves from xi = -1.00 cm to xf = 6.00 cm .

Part C

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

Part D

Determine the changeKf−Ki in the kinetic energy of the block as it moves from xi = -1.00 cm to xf = 6.00 cm .

Answer 1

A) Work done = F1.s

= 95* (0.06- - 0.01) = 6.67 J

B) Work done = F2.s = -45* (0.06--0.01) = - 3.15 J

C) total work = 6.67-3.15 = 3.52 J

D) change in KE = work done by all forces = 3.52 J