# A 2.40 kg snowball is fired from a cliff 7.69 m high. The snowball’s initial velocity is 13.0 m/s, directed 49.0° above the horizontal. (a)

Question

A 2.40 kg snowball is fired from a cliff 7.69 m high. The snowball’s initial velocity is 13.0 m/s, directed 49.0° above the horizontal. (a) How much work is done on the snowball by the gravitational force during its flight to the flat ground below the cliff? (b) What is the change in the gravitational potential energy of the snowball-Earth system during the flight? (c) If that gravitational potential energy is taken to be zero at the height of the cliff, what is its value when the snowball reaches the ground?

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1 year 2021-09-02T07:27:49+00:00 1 Answers 10 views 0

a) W = 180.87 J
, b)  ΔU = -180.87 J
, c)  ΔU = -180.87 J

Explanation:

a) Work is defined as

W = F .ds

Where bold indicates vectors, we can write the scalar product

W = F s cos θ

Where the angle is between force and displacement.

The force of gravity is the weight of the body, which is directed downwards and the displacement thickens the tip of the cliff at the bottom, so that it is directed downwards, therefore the angle is zero degrees

W = [tex]F_{g}[/tex] y

W = m g y

For this problem we must fix a reference system, from the statement it is established that the system is placed at the base of the cliff, so that final height is zero and the initial height (y₀ = 7.69m)

W = 2.40 9.8 (7.69-0)

W = 180.87 J

b) The potential energy is

U = mg y

The change in potential energy,

ΔU = [tex]U_{f}[/tex]- U₀

ΔU = mg ([tex]y_{f}[/tex]- y₀)

ΔU = 2.4 9.8 (0 -7.69)

ΔU = -180.87 J

c) in this case we change the reference system to the height of the cliffs, for this configuration

y₀ = 0

[tex]y_{f}[/tex] = -7.69 m

ΔU = 2.4 9.8 (-7.69 -0)

ΔU = -180.87 J