A steel ball of 0.18 kg is accidentally dropped 2.00 m above the ground. It ente

Question

A steel ball of 0.18 kg is accidentally dropped 2.00 m above the ground. It enters the soil, slows down and stops at a depth of 0.15m. Ignore air resistance during the drop.

a) give the formula for work and gravitational potential energy.

b) how much work was done by the soil to stop the ball?

c) Give the average force exerted on the ball by the soil.

d) what was the speed of the ball the moment it touches the soil?

e) if the ball had fallen in a frictionless curved drain pipe and left it vertically down, how far would it have entered the soil then?

Explanation / Answer

(a)

work E = mgh

gravitational potential energy = mgh

(c)

from the kinematic equation

v^2 - u^2 = 2as

v^2 - 0^2 = 2as

v = sqrt 2as = sqrt 2(9.8) (2.0 m) = 6.26 m/s

(b)

here v = u

W = 1/2 mv^2 - 1/2 mu^2 = 1/2 (0.18 kg) ( 6.26)^2 = 3.528 J

(c)

v^2 - u^2 = 2as

0- u^2 = 2as

a = -u^2/ 2s =- (6.26)^2/2 ( 0.15 m) = -130.62 m/s^2

average force

F = ma = 0.18 kg ( -130.62 m/s^2) =23.51 N

(d)

from the kinematic equation

v^2 - u^2 = 2as

v^2 - 0^2 = 2as

v = sqrt 2as = sqrt 2(9.8) (2.0 m) = 6.26 m/s

As per guidelines I worked first four parts of the problem

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