Two forces of magnitudes F1=60.00N and F2=45.0N act in opposite directions on a

Question

Two forces of magnitudes F1=60.00N and F2=45.0N act in opposite directions on a block, which sits atop a frictionless surface. Initially the center of the bock is at position x1=-1.00cm. at some later time, the block has moved to the right and its center is at a new position, xf=6.00cm

Find the work W1 done on the block by the force of magnitude F1 = 60.0N as the block moves from xi = -1.00cm to xf = 6.00cm .

Find the work W2 done by the force of magnitude F2 = 45.0N as the block moves from xi = -1.00cm to xf = 6.00cm .

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 = 6.00cm .

Explanation / Answer

a) work W1 done on the block by the force of magnitude F1 = 60.0N =
=60*(6-(-1))*10^-2
= 4.2 J

b) work W2 done by the force of magnitude F2 = 45.0N
= 45*(-7)*`10^-2
= -3.15 J

c)net work Wnet done on the block by the two forces = 4.20-3.15 = 1.05 J

d) changeKf?Ki in the kinetic energy of the block =net work Wnet done on the block by the two forces
= 1.05 J

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