Two forces, of magnitudes F1 = 90.0N and F2 = 35.0N , act in opposite directions

Question

Two forces, of magnitudes F1 = 90.0N and F2 = 35.0N , 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 = -6.00cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 2.00cm

A) Find the work W1 done on the block by the force of magnitude F1 = 90.0N as the block moves from xi = -6.00cm to xf = 2.00cm

B) Find the work W2 done by the force of magnitude F2 = 35.0N as the block moves from xi = -6.00cm to xf = 2.00cm

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

D) Determine the changeKf?Ki in the kinetic energy of the block as it moves from xi = -6.00cm to xf = 2.00cm .

Explanation / Answer

a) work done = force * displacement

so in work done by F1 = 90* (2-(-6)) * 10-2 = 7.20J

b) work done by F2 = -35* (2-(-6)) *10-2 = - 2.80J

c) net work done by two forces = 7.2J -2.8J = 4.4J

d)work done by the both forces on the block increases its total energy, as according to energy conservation. But here no change in potential is possible because of change in height is zero so change in kinetic energy = work done by F1 + work done by F2

change in kinetic energy Kf?Ki = 7.20-2.80 = 4.4 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.