A 1250 kg wrecking ball is lifted to a height of 12.7 m above itw resting point.

Question

A 1250 kg wrecking ball is lifted to a height of 12.7 m above itw resting point. When the wrecking ball is released, it swings toward an abandoned building and makes an indention of 43.7 cm in the wall.

a. What is the potential energy of the wrecking ball at a height of 12.7 m?

b. What is its kinetic energy at the time of striking the wall?

c. If the wrecking ball transfers all of its kinetic energy to the wall, how much force does the wrecking ball apply to the wall?

d. Why should a wrecking ball strike a wall at the lowest point in its swing?

Explanation / Answer

a)

Potential energy of the ball , PE = m*g*h

Potential energy of the ball , PE = 1250 * 9.8 * 12.7

Potential energy of the ball , PE = 155575 J

the Potential energy of the ball is 155575 J

b)
when the ball is released , the potential energy of the ball is converted to it's kinetic energy ,

hence , kinetic energy of the ball = 155575 J

c)

as stopping distance , d = 43.7 cm = 0.437 m

let the stopping force is F ,

using work energy theorum

F*0.437 = -155575

F = - 356000 N

the stopping force acting on the ball is - 356000 N

d)

as the kinetic energy of the ball is maximum at the lowest point ,

it exerts maximum force at the lowest point on the object it is stricking

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