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

Question

Two forces, of magnitudes F1 = 65.0 N and F2 = 50.0 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 = 4.00 cm .

Part A

Find the work W1 done on the block by the force of magnitude F1 = 65.0 N as the block moves from xi = -3.00 cm to xf = 4.00 cm .

Part B

Find the work W2 done by the force of magnitude F2 = 50.0 N as the block moves from xi = -3.00 cm to xf = 4.00 cm .

Part C

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

Part D

Determine the changeKfKi in the kinetic energy of the block as it moves from xi = -3.00 cm to xf = 4.00 cm .

Explanation / Answer

here,

A)

the work doe by F1 , W1 = F1 * ( 0.04 - (- 0.03)) J

W1 = 65 * ( 0.07) J = 4.55 J

B)

the work doe by F2 , W2 = F2 * ( 0.04 - (- 0.03)) J

W2 = - 50 * ( 0.07) J = - 3.5 J

C)

using work energy theorm

the net work done , W = change in KE

change in KE = 4.55 - 3.5 = 1.05 J

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