A1.50-kg book is sliding along a rough horizontal surface. at point a it is moving at 3.21 m> s, and at point b it has slowed to 1.25 m> s. (a) how much work was done on the book between a and b? (b) if -0.750 j of work is done on the book from b to c, how fast is it moving at point c? (c) how fast would it be moving at c if +0.750 j of work was done on it from b to c?

Given:
Mass, m = 1.50 kg
Velocity at point a, v1 = 3.21 m/s
Velocity at point b, v2 = 1.25 m/s

Part (a)
Because the velocity decreased, work was done to slow down the book. The work is equal to the loss in kinetic energy.
Work done = (1/2)*(1.5 kg)*(3.21² - 1.25² m²/s²) = 6.556 J

-6.556 j

Part (b)
Let v3 =  velocity of the book at point c.
Work done is -0.75 J from point b to point c.
This will cause loss of kinetic energy of the book.
Therefore
(1/2)*(1.5 kg)*(1.25² - v3² m²/s²) = 0.75 J
0.75(1.5625 - v3²) = 0.75
1.5625  - v3² = 1
v3² = 0.5625
v3 = 0.75 m/s

The velocity at c is 0.75 m/s

Part (c)
If +0.75 J of work is performed on the book between b and c, the kinetic energy of the book will increase. Therefore
(1/2)*(1.5 kg)*(v3² - 1.25² m²/s²) = 0.75 J
0.75(v3² - 1.5625) = 0.75
v3² - 1.5625 = 1
v3² = 2.5625
v3 = 1.6 m/s

The velocity at c is 1.6 m/s

10 neutrons is the answer

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