Two forces, of magnitudes F 1 = 95.0 N and F 2 = 30.0 N , act in opposite direct

Question

Two forces, of magnitudes F1 = 95.0 N and F2 = 30.0 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 = -3.00 cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 6.00 cm .

a.Find the work W1 done on the block by the force of magnitude F1 = 95.0 N as the block moves from xi = -3.00 cm to xf = 6.00 cm .

Express your answer numerically, in joules.

b.Find the work W2 done by the force of magnitude F2 = 30.0 N as the block moves from xi = -3.00 cm to xf = 6.00 cm .

Express your answer numerically, in joules.

c.Part C What is the net work Wnet done on the block by the two forces?

Express your answer numerically, in joules.

D.Part D Determine the changeKfKi in the kinetic energy of the block as it moves from xi = -3.00 cm to xf = 6.00 cm .

Express your answer numerically, in joules.

Explanation / Answer

The net displacement =xf-xi = 9cm = 0.09m

1. Work done by F1 = force x displacement = 95 x 0.09 = 8.55 Joules

Hence the work done by F1 is 8.55 joules.

2. Work done by F2 = force x displacement = -30 x 0.09 = -2.7 Joules

Hence the work done by F2 is -2.7 Joules

3. Wnet = F1 - F2 = 95 - 30 = 65N

Work done by Wnet = 65 x 0.09 = 5.85 Joules

Hence the work done by net force = 5.85 Joules

4. From work energy theorem

Change in kinetic energy = net work done

Kf - ki = net work done = 5.85 Joules

Hence change in kinetic energy = 5.85 Joules.

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