Question

A 1.50 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 .

Part A

How much work was done on the book between A and B ?

Part B

If -0.750J 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 0.750J of work were done on it from B to C ?

Answer 1

Mass of the book = m = 1.5 kg

Speed of the book at point A = V₁ = 3.21 m/s

Speed of the book at point B = V₂ = 1.25 m/s

Work done on the book between A and B = W₁

The work done on the book between A and B is equal to the change in kinetic energy of the book.

W₁ = mV₂²/2 - mV₁²/2

W₁ = (1.5)(1.25)²/2 - (1.5)(3.21)²/2

W₁ = -6.556 J

Speed of the book at point C if -0.75 J of work is done on the book from B to C = V₃

Work done on the book from B to C = W₂ = -0.75 N

W₂ = mV₃²/2 - mV₂²/2

-0.75 = (1.5)V₃²/2 - (1.5)(1.25)²/2

V₃ = 0.75 m/s

Speed of the book at point C if 0.75 J of work is done on the book from B to C = V₄

New work done on the book from B to C = W₃ = 0.75 J

W₃ = mV₄²/2 - mV₂²/2

0.75 = (1.5)V₄²/2 - (1.5)(1.25)²/2

V₄ = 1.6 m/s

A) Work done on the book between A and B = -6.556 J

B) Speed of the book at point C if -0.75 J of work is done on the book from B to C = 0.75 m/s

C) Speed of the book at point C if 0.75 J of work is done on the book from B to C = 1.6 m/s