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A 3.2 kgsloth hangs 3.0 mabove the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point y=0to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

Short Answer

Expert verified
  1. The gravitational potential energy U is,U=94j
  2. The kinetic energy K is,K=94J
  3. The speed v of the sloth just before it reaches the ground is,h=7.7m/s

Step by step solution

01

Listing the given quantities

The mass of the sloth hanging is,m=3.2kg

The height of the sloth hanging from the ground is h=3.0m.

02

Understanding the concept of potential and kinetic energy  

By using the equation for gravitational potential U, we can find thegravitational potential energy U. Here, mechanical energy is conserved. So, kinetic energy Kis equal to the gravitational potential energy U. Finally, by using the equation for kinetic energy K, we can find thespeedvof the sloth just before it reachesthe ground.

Formula:

The gravitational potential is,U=mgh

The kinetic energy is,K=12mv2

03

Step 3(a): Calculation of gravitational potential energy U

U=mgh=3.2×9.8×3.0=94J

Hence, the gravitational potential energy is, U =94 J

04

Step 4(b): Calculation of kinetic energy

The mechanical energy is conserved; hence, we get the kinetic energyKis

K=UK=94J

Hence, the kinetic energy is,K=94 J

05

Step 5(c): Calculation the speed v of the sloth just before it reaches the ground

We know that kinetic energy is given by

K=12mv2v2=2Km

Therefore, we get the speed v as

v=2Km=2×943.2=7.7m/s

Hence, the speed of the sloth just before it reaches the ground is,v=7.7 m/s

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