Question #145580
A 30-kg rock is on top of a 50-m cliff. How much of its potential energy is lost when it falls and
reaches 20 m above the ground
1
Expert's answer
2020-11-23T10:28:24-0500

The loss om potential energy is:


ΔU=U2U1\Delta U = |U_2 - U_1|

Where U1U_1 is the initial energy and U2U_2 is the final potential energy of the rock. The initial is:


U1=mgh1U_1 = mgh_1

where m=30kgm = 30kg is the mass of the rock, g=9.81m/s2g = 9.81m/s^2 is the gravitational acceleration, and h1=50mh_1 = 50m is the initial height. Thus, obtain:


U1=309.8150=14715JU_1 = 30\cdot 9.81\cdot 50 = 14715J

Similarly, the final potential energy is:


U2=mgh2U_2 = mgh_2

where h2=20mh_2 = 20m is the final height of the rock. Thus, get:


U2=309.8120=5886JU_2 = 30\cdot 9.81\cdot 20 = 5886J

Then the loss is:


ΔU=5886J14715J=8829J\Delta U = |5886J-14715J| = 8829J

Answer. 8829 J.


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