Two forces, of magnitude F1 = 80.0 N and F2 = 45.0 N, act in opposite directions

Question

Two forces, of magnitude F1 = 80.0 N and F2 = 45.0 N, act in opposite directions on a block which sits atop a frictionless surface. Initially, the center of the block is at position xi = -1.00 cm. At some latertime, the block has moved to the right, and its center is at a new position, xf = 2.00 cm.

(a) Find the work W1 done on the block by the force of magnitude F1= 80.0 N as the block moves from xi = -1.00 cm to xf = 2.00 cm.

(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 = 2.00 cm.

(c) What is the net work, W_net, done on the block by the two forces?

(d) Determine the change, K_f - K_i, in the kinetic energy of the block as it moves from xi = -1.00 cm to xf = 2.00 cm.

Explanation / Answer

(a) work done = force*displacement

=> W1 = 80*0.03 = 2.4 J

(b) W2 = F2*d

=> W2 = 45*0.03 = 1.35 J

(c) net work done = W1 - W2

so W_net = 2.4 - 1.35 = 1.05 J

(d) by kinetic energy work theorem,

change in kinetic energy = net work done

=> K_f - K_i = W_net

so K_f - K_i = 1.05 J

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