Two forces, of magnitudes F 1 = 85.0N and F 2 = 40.0N , act in opposite directio

Question

Two forces, of magnitudes F1 = 85.0N and F2 = 40.0N , 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 = -3.00cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 4.00cm .

Part A:

Find the work W1 done on the block by the force of magnitude F1 = 85.0N as the block moves from xi = -3.00cm to xf = 4.00cm .

Express your answer numerically, in joules.

Part B:

Find the work W2 done by the force of magnitude F2 = 40.0N as the block moves from xi = -3.00cm to xf = 4.00cm .

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 = -3.00cm to xf = 4.00cm .

Express your answer numerically, in joules.

Two forces, of magnitudes F1 = 85.0N and F2 = 40.0N , 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 = -3.00cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 4.00cm . Part A: Find the work W1 done on the block by the force of magnitude F1 = 85.0N as the block moves from xi = -3.00cm to xf = 4.00cm . Express your answer numerically, in joules. Part B: Find the work W2 done by the force of magnitude F2 = 40.0N as the block moves from xi = -3.00cm to xf = 4.00cm . 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 = -3.00cm to xf = 4.00cm . Express your answer numerically, in joules.

Explanation / Answer

A)
work done by a force = force * dispalcement * cos(theta)

work done by force F1 = (0.04 - (-0.03))*85

work done by force F1 = 5.95 J

B)

here , for work done by force by F2 = (xf - xi) * F2 * cos(180)

work done by force by F2 = (0.04 + 0.03) * 40 *cos(180)

work done by force by F2 = -2.8 J

C)

Net work done = WF1 + WF2

Net work done = 5.95 - 2.8

Net work done = 3.15 J

D)

Using work energy theorum

change in kinetic energy , Kf - Ki = Net work done

change in kinetic energy , Kf - Ki = 3.15 J

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