Two forces, of magnitudes F 1 = 65.0N and F 2 = 50.0N , act in opposite directio

Question

Two forces, of magnitudes F1 = 65.0N and F2 = 50.0N , 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 = -6.00cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 2.00cm .

Part A

Find the work W1 done on the block by the force of magnitude F1 = 65.0N as the block moves from xi = -6.00cm to xf = 2.00cm .

Express your answer numerically, in joules.

Part B

Find the work W2 done by the force of magnitude F2 = 50.0N as the block moves from xi = -6.00cm to xf = 2.00cm .

Express your answer numerically, in joules.

Part C

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

Express your answer numerically, in joules.

Part D

Determine the changeKf?Ki in the kinetic energy of the block as it moves from xi = -6.00cm to xf = 2.00cm .

Express your answer numerically, in joules.

Explanation / Answer

part A

w= F.S cos(a)

a=0, F=65 N S=8 cm

w1 = + 5.20 J

part B:

w2 = - 50 *0.0 8 = - 4.0 J

Part c:

net work done, Wnet = 5.2-4 = 1.2 J

Part D;

So the change in kinetic energy is equal to the net work done.

hence , Kf -Ki = 1.2 J

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