A 1.45 kg book is sliding along a rough horizontal surface. At point A it is moving at 3.25 m/s , and at point B it has slowed to 1.34 m/s .

Part A
How much work was done on the book between A and B?
Part B
If -0.750 J of work is done on the book from B to C, how fast is it moving at point C?
Part C
How fast would it be moving at C if 0.750 J of work were done on it from B to C?

Part A:

To find the work done on the book between points A and B, we can use the work-energy principle. The work-energy principle states that the net work done on an object is equal to its change in kinetic energy.

The formula for work done is given by:
Work = Force * Distance * cos(theta)

However, in this case, we are dealing with a book sliding on a rough horizontal surface, so there is friction involved. Friction is a force that acts against the motion and does negative work. Therefore, the formula becomes:
Work = Force_applied * Distance * cos(theta) - Force_friction * Distance * cos(theta)

Since the book is sliding horizontally, the angle between the applied force and displacement is 0 degrees, so cos(0)=1. Therefore, the cos(theta) term is not really needed, but we can include it for completeness.

The work done by friction can be calculated using the formula:
Force_friction = coefficient_friction * Normal_force

Given that we have no information about the coefficient of friction or the normal force, we cannot directly determine the work done by friction. However, we are given the initial and final velocities of the book, which can help us find the work done on the book.

To find the net work done on the book, we can use the equation:
Work_net = Change_in_kinetic_energy

The change in kinetic energy is given by:
Change_in_kinetic_energy = (1/2) * m * (v_f^2 - v_i^2)

where m is the mass of the book, v_f is the final velocity, and v_i is the initial velocity.

Plugging in the given values:
m = 1.45 kg,
v_i = 3.25 m/s,
v_f = 1.34 m/s,

we can calculate the work done on the book between points A and B:
Work_net = (1/2) * 1.45 kg * ((1.34 m/s)^2 - (3.25 m/s)^2)

Part B:
To find the final velocity of the book at point C, we need to consider the work done on the book from B to C.

Given that -0.750 J of work is done on the book, we can use the work-energy principle again to relate the work done to the change in kinetic energy.

Using the equation:
Work_done = Change_in_kinetic_energy

We can rearrange the equation to solve for the final velocity:
Change_in_kinetic_energy = (1/2) * m * (v_f^2 - v_i^2)

Given that the work done is -0.750 J, we have:
-0.750 J = (1/2) * 1.45 kg * (v_f^2 - (1.34 m/s)^2)

Solving for v_f, we can calculate the final velocity of the book at point C.

Part C:
To find the final velocity of the book at point C, if 0.750 J of work is done on it from B to C, we can apply the same approach as in Part B.

Given that the work done is 0.750 J, we have:
0.750 J = (1/2) * 1.45 kg * (v_f^2 - (1.34 m/s)^2)

Solving for v_f, we can calculate the final velocity of the book at point C.

Ke at A = (1/2)(1.45)(3.25)^2 = 7.66 Joules

Ke at B = (1/2)(1.45)(1.34)^2 = 1.30 Joules
so work done= delta ke = 1.3-7.66 = -6.36 J

B) Now you should be able to continue.
slows during B, speeds up during C