From Chapter 6: Work and Energy A steel ball of 0.18 kg is dropped 2.00 m above

Question

From Chapter 6: Work and Energy

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

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

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) 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)

consider the free fall motion of the ball

Vi = initial velocity at the top = 0 m/s

Vf = final velocity just before hitting the soil

a = acceleration = 9.8 m/s2

Y = displacement = 2 m

Using the equation

Vf2 = Vi2 + 2 a Y

Vf2 = 02 + 2 (9.8) (2)

Vf = 6.26 m/s

b)

Using conservation of energy

kinetic enery + Potential just before hitting the soil = work done by soil to stop the ball

W = mgh + (0.5) m V2

W = 0.18 (9.8) (2) + (0.5) (0.18) (6.26)2

W = 7.055 J

c)

Vi = initial velocity just before hitting the soil = 6.26 m/s

Vf = final velocity = 0 m/s

d = stopping distance = 0.15 m

Vf2 = Vi2 + 2 a' d

02 = 6.262 + 2 a' (0.15)

a' = -130.625 m/s2

average force = ma' = 0.18 x 130.625 = 23.51 N

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