A 1.60 kg snowball is fired from a cliff 8.93 m high. The snowball's initial velocity is 12.4 m/s, directed 30.0° 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?

a)140.0224 J b) - 140.0224 J c)- 140.0224 J

Explanation:

Work done is defined as force acting in a direction multiplied by the distance traveled in that direction. It is defined by the formula

W = F cos θ

a) the amount of work done by the gravitational force during its flight to the flat ground = mgh cosθ where θ = 0 since it is traveling downward

W = 1.60 × 8.93 × 9.8 cos0 = 140.0224 J

b) The change in the potential energy = final potential energy - initial potential energy = 0 - 140.0224 J = - 140.0224 J

c) taken  gravitational potential energy to be zero at the height and since the snow ball is travelling below it reference point, then

the value when its reached the ground = -mgh = - 140.0224 J

(A)Workdone –140.022J (B) ΔPE –140.022J (C)

PE2 = –140.022J

Explanation:

The workdone by the gravitational force is given by the product of the force if gravity acting on the mass and the vertical displacement of the mass.

The for is the weight of the body.

W = mg

The vertical displacement = h

Workdone = W ×h = mgh

m = 1.60kg

g = 9.8m/s²

h = ‐8.93m

Workdone = –1.60×9.8×8.93 = – 140.022J

(b) From the work and energy theorem, which states that, the change in potential energy is the total workdone in displacing it vertically over a distance h.

So the change in potential energy, ΔPE = Workdone = –140.022J

(C)

Taking the top of the cliff as PE1 and the ground as PE2

PE2 – PE1 = workdone

Given that PE1 = 0J

So substituting the values in the equation above.

PE2 – 0 = –140.022J.

PE2 = –140.022J

In the whole of this calculation we did not need to use the velocity given or calculate the height the snowball rises during the course of its motion as the workdone in rising that height against gravity will be cancelled out by the force of gravity during descending part of the motion of the snowball as the force of gravity will do an equal but opposite (negative) work on the snowball. So for this kind of problem only the start and end points of the motion are important for consideration as every other point is traveled twice vertically. First going up and second when coming back down.