Physic helppp plz?

Question

1.a physics student lifts her 2kg pet rock 2.8m straight up. He then lets it drop to the ground. Use the Law of Conservation of Mechanical Energy to calculate how fast the rock will be moving (a)half way down and (b)just before it hits the ground
2.A 65kg girl is running with a speed of 2.5m/s. How much kinetic energy does she have? She grabs on to a rope which is hanging from the ceiling, and swings from the end of the rope. How high off the ground will she swing?
3.How much kinetic energy will the 80kg skier sliding down the frictionless slope have when he is 2/3 of the way down the ramp? The vertical height of the ramp is 60m.
4.A rubber ball falls from a height of 2m, bounces off the floor and goes back up to a height of 1.6m. What percentage of its initial gravitational potential energy has been lost? Where does this energy go? Has the Law of Conservation been violated?
5.How much work must be done to increase the speed of a 12kg bike ridden by a 60kg from 8.2m/s to 12.7m/s?

Answer 1

1. m = 2 kg
g= 9.8 m/s^2
h = 2.8 m

PE = mgh = 54.88 J

KE at the initial point = 0 since the velocity is 0.

So total Mechanical energy = PE + KE = 54.88 J

(a) h' = 1.4 m
So PE = 2 * 9.8 * 1.4 = 27.44 J

ME = PE + KE
KE = 27.44 J

1/2 mv^2 = 27.44 J
v^2 = 27.44
v = 5.24 m/s

(b) h'' = 0 m
PE = 0

So KE = 54.88 J
1/2 mv^2 = 54.88 J
v = 7.41 m/s

2. KE = mv^2 / 2 = 65 * 2.5 * 2.5 / 2 = 203.125 J
PE = 0 since the height is 0.

ME = 203.125 J

At the highest point:
KE = 0 since the velocity has become 0.
So PE = 203.125 J
mgh = 203.125 J
h = 0.32 m = 32 cm

3. At the top of the slope:
h = 60 m
PE = 47040 J
KE = 0
ME = 47040 J

At the point when
h' = 2/3 * 60 = 40 m
PE = 31360 J
So KE = 15680 J

1/2 mv^2 = 15680 J
v = 19.8 m/s

4. Initial ME = PE + KE = (2 * M * g) + 0 = 2 Mg
Final ME = PE + KE = (1.6 * M * g) + 0 = 1.6 Mg

Energy lost = 0.4 Mg

%age of energy lost = 0.4Mg / 2Mg * 100 = 20 %

This energy has gone into the Earth. The law of conservation has not been violated however the acceleration produced in the earth is negligible due to its huge mass.

5. Total mass = 12 + 60 = 72 kg
u = 8.2 m/s
v = 12.7 m/s
Work done = Change in KE = 1/2 mv^2 - 1/2 mu^2
= m/2 (v^2 - u^2) = 3385.8 J

Hope this helps.

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Answer 2

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Answer 3

It is not worth the time to answer 5 questions to get points. Give us a break !!