Two forces, of magnitudes F 1 = 80.0 N and F 2 = 40.0 N , act in opposite direct

Question

Two forces, of magnitudes F1 = 80.0N 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 = -5.00cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 3.00cm .

Find the work W1 done on the block by the force of magnitude F1 = 80.0N as the block moves from xi = -5.00cm to xf = 3.00cm .

Find the work W2 done by the force of magnitude F2 = 40.0N as the block moves from xi = -5.00cm to xf = 3.00cm .

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

Determine the changeKf?Ki in the kinetic energy of the block as it moves from xi = -5.00cm to xf = 3.00cm .

Explanation / Answer

1) W1 = F1*(xf-xi)*cos(0) = 80*(0.03-(-0.05))*(1) = 6.4 J

2) W2 = F2*(xf-xi)*cos(180) = 40*(0.03-(-0.05))*(-1) = -3.2 J

3) Wnet = W1 + W2 = +3.2 J

4) Wnet = Kf - Ki = +3.2 J

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