Two forces, of magnitudes F1= 80.0N and F2= 50.0N, act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position x1= -3.00cm. At some later time, the block has moved to the right, and its center is at a new position, x2= 2.00cm.

a)Find the work W1 done on the block by the force of magnitude F1= 80.0N as the block moves from x1= -3.00 cm to x2= 2.00 cm
b)Find the work done by the force of magnitude F2 = 50.0 N as the block moves from x1= -3.00 Ncm to x2= 2.00cm.
c)What is the net work Wnet done on the block by the two forces?
d)Determine the change in the kinetic energy (Kf-Ki) of the block as it moves from x1= -3.00 cm to x2= 2.00 cm.

a) F_1=80*(0.03+0.02)

b) F_2=50*(0.03+0.02)
c) answer part a + answer part b
d) same as part c

a) Well, isn't it fascinating that the force F1 is playing a game of tug-of-war with the block? To find the work done by F1, we can use the equation W1 = F1 * d1, where d1 is the displacement. In this case, the block moved from x1 = -3.00 cm to x2 = 2.00 cm, which gives us a displacement of d1 = 5.00 cm. But wait a minute, we need to convert that to meters, because, you know, the metric system is the cool kid on the block. So, d1 = 0.050 m. Now we can calculate the work W1 done on the block by clowning around with the equation: W1 = 80.0 N * 0.050 m. And voila, W1 turns out to be 4.00 J. Ta-da!

b) And now, it's time to give F2 a chance to shine in the work show! To find the work done by F2, we can use the same equation as before: W2 = F2 * d2. Again, the block moved from x1 = -3.00 cm to x2 = 2.00 cm, so the displacement is d2 = 5.00 cm = 0.050 m. Now let's calculate the work W2: W2 = 50.0 N * 0.050 m. And look at that, W2 turns out to be 2.50 J. Applause, please!

c) Ah, now it's time to add some drama to the mix. The net work Wnet done on the block by the two forces is as simple as adding W1 and W2. So, Wnet = W1 + W2 = 4.00 J + 2.50 J = 6.50 J. And oh boy, the audience goes wild!

d) Now it's time for the grand finale! To find the change in kinetic energy (Kf - Ki), we need to subtract the initial kinetic energy from the final kinetic energy. But since there's no mention of any initial or final velocities here, we can assume that the initial and final velocities are zero. Therefore, the initial kinetic energy Ki is zero, and the final kinetic energy Kf is also zero. So the change in kinetic energy (Kf - Ki) = 0 J. Oh, the suspense was unbearable, wasn't it?

To find the work done by a force, you can use the formula:

Work = Force * Distance * Cosine(θ)

Where:
- Force is the magnitude of the force in Newtons (N)
- Distance is the distance over which the force is applied in meters (m)
- θ is the angle between the force and the direction of motion (in this case, the angle is 180 degrees because the forces are acting in opposite directions)

a) To find the work done by force F1 as the block moves from x1 = -3.00 cm to x2 = 2.00 cm:
- The magnitude of force F1 is given as 80.0 N.
- The distance over which the block moves is the difference between x2 and x1, which is 2.00 cm - (-3.00 cm) = 5.00 cm = 0.05 m.
- The angle between the force and the direction of motion is 180 degrees.

Using the formula, we can calculate the work done by force F1:

Work1 = F1 * Distance * Cosine(180 degrees)
= 80.0 N * 0.05 m * Cosine(180 degrees)
= -4.00 J

Therefore, the work done by force F1 is -4.00 Joules.

b) To find the work done by force F2 as the block moves from x1 = -3.00 cm to x2 = 2.00 cm:
- The magnitude of force F2 is given as 50.0 N.
- The distance over which the block moves is still 0.05 m.
- The angle between the force and the direction of motion is still 180 degrees.

Using the formula, we can calculate the work done by force F2:

Work2 = F2 * Distance * Cosine(180 degrees)
= 50.0 N * 0.05 m * Cosine(180 degrees)
= -2.50 J

Therefore, the work done by force F2 is -2.50 Joules.

c) The net work done on the block by the two forces can be calculated by summing the work done by each force:

Wnet = Work1 + Work2
= -4.00 J + (-2.50 J)
= -6.50 J

Therefore, the net work done on the block by the two forces is -6.50 Joules.

d) The change in kinetic energy (Kf - Ki) can be determined using the work-energy principle, which states that the net work done on an object equals the change in its kinetic energy.

Kf - Ki = Wnet

Therefore, the change in kinetic energy can be calculated as:

Kf - Ki = -6.50 J

Note: Since the problem does not provide initial or final velocities of the block, it is not possible to determine the actual change in kinetic energy. The given information only allows us to calculate the net work done on the block.

a) 4

You don't say which force acts in the direction of motion and which does not. Presumably F1, which is larger, is along the direction of motion. Otherwise the block would move backwards.

One force will do positive work and the other, negative work.

For (c), add them.

For (d) Kf - Ki equals the net work done (since there is no friction)

Do not expect step by step solutions here. We prefer to provide quidelines so you can do the work yourself.