Question

A 1.80 kg snowball is fired from a cliff 9.10 m high. The snowball's initial velocity is 17.4 m/s, directed 44.0° 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?

Answer 1

a) The work done is 160.52 J

b) The change in the gravitational potential energy on the snowball is -160.52 J

c) The gravitational potential energy that is taken to be zero at the height of the cliff is -160.52 J

Step-by-step explanation:

Given:

m = 1.8 kg

h = 9.1 m

vi = 17.4 m/s

θ = 44°

a) The work done on the snowball is:

`W_(g) =Fdcos\theta =mghcos\theta`

Here θ = 0 because they are in the same direction

`W_(g) =1.8*9.8*9.1*cos0=160.52J`

b) The change in the gravitational potential energy on the snowball is:

ΔU = -Wg = -160.52 J

c) The gravitational potential energy that is taken is:

`U_(g) =-mgh=-1.8*9.8*9.1=-160.52J`