Determine the changeKf-Ki in the kinetic energy of the block as it moves from xi = - 4.00 cm to xf = 6.00 cm.

A) W1 = 10.0 J

B) W2 = - 2.00 J

C) Wnet = 8.0 J

D) kf - ki = 8.0 J

Explanation:

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Two forces, of magnitudes F1 = 100 N and F2 = 20.0 N, act in opposite directions on a block, which sits atop a friction-less surface. Initially, the center of the block is at position xi = - 4.00 cm. At some later time, the block has moved to the right, and its center is at a new position, xf = 6.00 cm.

A) Find the work W1 done on the block by the force of magnitude F1 = 100 N as the block moves from xi = - 4.00 cm to xf = 6.00 cm.

B) Find the work W2 done by the force of magnitude F2 = 20.0 N as the block moves from xi = - 4.00 cm to xf = 6.00 cm.

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

D) Determine the change Kf-Ki in the kinetic energy of the block as it moves from xi = - 4.00 cm to xf = 6.00 cm.

A) The equation of work is the following:

W = F · d

Where:

W = work.

F = applied force.

d = traveled distance.

The traveled distance from - 4.00 cm to 6.00 is 10.00 cm (final position - initial position = 6.00 cm - (-4.00 cm) = 10.00 cm = 0.1000 m).

Then, the work done by F1 is the following:

W1 = F1 · d

W1 = 100 N · 0.1000 m = 10.0 J

B) Similarly, we can calculate the work done by the force F2:

W2 = F2 · d

Notice that since the force is applied in the direction opposite to the displacement, the work is negative:

W2 = - 20.0 N · 0.100 m = - 2.00 J

C) The net work can be calculated by adding the work done by all the forces on the object. In this case:

Wnet = W1 + W2

Wnet = 10.0 J - 2.00 J = 8.0 J

D) According to the work-energy theorem, the net work on an object is equal to its change in kinetic energy:

Wnet = ΔKE = kf - ki = 8.0 J

This means that the total work done on the block increased its kinetic energy by 8.0 J.