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3 years ago

8

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 t

o 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 ?

1 answer:

garik1379 [7]3 years ago

3 0

I assume friction is the only force acting on the book as it slides.

(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:

<em>W</em> = ∆<em>K</em>

<em>W</em> = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²

<em>W</em> ≈ -6.56 J

(B) Using the work-energy theorem again, the speed <em>v</em> of the book at point C is such that

-0.750 J = 1/2 (1.50 kg) <em>v</em> ² - 1/2 (1.50 kg) (1.25 m/s)²

==> <em>v</em> = 0.750 m/s

(C) Take the left side to be positive, then solve again for <em>v</em>.

0.750 J = 1/2 (1.50 kg) <em>v</em> ² - 1/2 (1.50 kg) (1.25 m/s)²

==> <em>v</em> ≈ 1.60 m/s

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$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

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