You drop a 2.00 kg textbook out of a window 10.0 m above the ground. A friend ca

Question

You drop a 2.00 kg textbook out of a window 10.0 m above the ground. A friend catches
1.50 m above the ground

a) How much work is done on the book by its weight as it drops?

b) What is the change in gravitational potential energy of the book during the drop?

If gravitational potential energy is 0 at ground level, what is the potential energy
c) at the moment you drop the book and
d) at the moment the book is caught

f) what is the speed of the textbook when it reaches the hands?

g) If we used a second textbook with twice the mass, what would its speed be?

Explanation / Answer

Let h2 = 10.0 m  and h2 = 1.50 m

a) Here work done = change in potential energy = mg(h2-h1) = 2*9.8*(10-1.50) = 166.6 J

b) change in potential energy = work done = mg(h2-h1) = 2*9.8*(10-1.50) = 166.6 J

c) Potential energy = mgh2 =2*9.8*10 = 196 J

d) Potential energy = mgh2 = 2*9.8*1.5 = 29.4 J

f) V = 2g(10-1.5) = 12.9 m/sec

g) As V is not depend upon mass,so V remains = 12.9 m/sec

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