Someone drops a 50 − g pebble off of a docked cruise ship, 70.0 m from the water line. a person on a dock 3.0 m from the water line holds out a net to catch the pebble. (a) how much work is done on the pebble by gravity during the drop? (b) what is the change in the gravitational potential energy during the drop? if the gravitational potential energy is zero at the water line, what is the gravitational potential energy (c) when the pebble is dropped? (d) when it reaches the net? what if the gravitational potential energy was 30.0 joules at water level? (e) find the answers to the same questions in (c) and (d).

a) The work done on the pebble is 32.9 J.

b) The change in the gravitational potential energy is -32.9 J.

c) When the pebble is dropped, the potential energy is 34.3 J.

d) When the pebble reaches the net, the potential energy is 1.47 J.

e) If the potential energy is 30.0 J at water level, it will be 31.47 J at the net and 64.3 J at 70.0 m.

Explanation:

a) The work done by the gravity force can be calculated using the following equation:

W = F · Δy

Where:

W = work

F = gravity force

Δy = vertical displacement (final height - initial height)

The gravity force can be calculated as:

F = m · g

Where:

F = gravity force

m = mass

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Then, the work done on the pebble by the gravity force will be:

W = m · g · (hf - hi) (hf = final height, hi = initial height)

W = 0.050 kg · (-9.81 m/s²) · (3.0 m - 70.0 m)

W = 32.9 J

The work done on the pebble is 32.9 J.

(Notice that the work is positive. That means that the force that does the work is in the same direction as the movement).

b) The potential energy (EP) can be calculated as follows:

EP = m · g · h (h = height)

The change in the gravitational potential energy is calculated as the difference of the potential energy between the two positions:

ΔEP = EPf - EPi

Where:

ΔEP = change in the gravitational potential energy.

EPf = final potential energy.

EPi = initial potential energy.

Then:

ΔEP = EPf - EPi

ΔEP = (0.050 kg · 9.81 m/s² · 3.0 m) - (0.050 kg · 9.81 m/s² · 70.0 m)

ΔEP = -32.9 J

c) When the pebbel is dropped, the potential energy will be:

EP = m · g · h

EP = 0.050 kg · 9.81 m/s² · 70.0 m = 34.3 J

d) When it reaches the net h = 3.0 m. Then:

EP = 0.050 kg · 9.81 m/s² · 3.0 m = 1.47 J

e) We have to add 30 J to the potential energy values calculated above:

The potential energy at 70.0 m will be: 34.3 J + 30 J = 64.3 J

The potential energy at 3.0 m will be: 1.47 J + 30 J = 31.47 J

that would be b i just took the test : )

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s