A woman of 60kg mass runs up 7 flights of stairs, her initial displacement vecto

Question

A woman of 60kg mass runs up 7 flights of stairs, her initial displacement vector is d= 2.25mi, at a constant rate of 2m/s, her final displacement vector is d=1mi + 21 mj. What is her gravitational potential energy with regards to her starting position? The next day she starts with the same final displacement vector and run the same 7 flights of stairs to the same final displacement vector. But she runs at constant rate of 4 m/s. What is her potential energy? On the third day she ran again, but this time her final displacement vector was given by d=150mi + 21mj. What is her potential energy? If she were then to jump out the window of the building on the 7th story, what would her velocity be just before she landed in the water of the pool below?

Explanation / Answer

In all the three cases her final potential energy will be same as the heoght attained by her is constant 21m in all three of the cases.

Potentail Energy = m g h = 60*9.8 *21 = 12.348 kJ

If she jumps from this height to the initial height

All her potential enrgy will get converted to kinetic energy.

PE = KE

1/2 m V^2 = 12348

v = 20.28 m/s

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