Two forces, of magnitudes F 1 = 75.0 N and F 2 = 30.0 N , act in opposite direct

Question

Two forces, of magnitudes F1 = 75.0N  and F2 = 30.0N , 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 = -1.00cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 3.00cm .

Two forces, of magnitudes F1 = 75.0N and F2 = 30.0N , 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 = -1.00cm . 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 = 75.0N as the block moves from xi = -1.00cm to xf = 3.00cm . Find the work W2 done by the force of magnitude F2 = 30.0N as the block moves from xi = -1.00cm 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 = -1.00cm to xf = 3.00cm .

Explanation / Answer

part 1) W1 = F1*(xf - xi)

= 75*(.03-(-.01)) = 3 joules

part 2) W2 = F2 *(xf-xi)

this work will be negative because force and displace ment direction are antiparallel

= -30*(.04)

= -1.2 joules

part 3) W(net) = W1 +W2 = 3-1.2 = 2.8 joules

part 4) change in kinetic energy = kf- ki

this change should be eqaul to total work done

so 2.8 joules = kf - li

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