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Several Energy Related Questions

1. Homework Statement
1) If a planet moves in a circle about the sun, how much work does the force of gravity do on the earth? Give reasons for your answer.

2) A force of 120N pushes a crate mass 15kg along the ground at a constant speed of 3.2 m/s for a distance of 6.2m. What amount of heat energy was generated at this time?

3)The earth has a mass of 5.98 x10^24 kg and a speed of 2.97 m/s. The sun, which is 1.49 x10^11m away, exerts 3.56 X10^11 N on it. If the earth orbited the sun in a perfect circle, how much work would the sun do on the earth in one second?

4) A car of mass 1350kg rolls from rest down a hill 235m long and 25m high. If its speed at the bottom of the hill is 12.0m/s, what's the force of friction on the car?


2. Homework Equations
Fnet=ma
pi*diameter
Ep=mgh
Ek=(0.5)mv^2

3. The Attempt at a Solution
1) I said none because the disatnce in the equation force times distance has to be a parallel distance to the force exerted on it, and its not in this case.

2)6.2m/3.2m/s=1.9375 sec
120N-Ff=m(0)
Ff=120N
W=Fd
W=120x6.2
W=744J

3)2(1.49 x10^11m)pi
d=2.97m
(3.56 x10^11 N)(2.97m)=1.06 x10^12 J

4)(1350kg)(9.8)(25m)
Ep'=330750 J
Ek2=(0.5)1350(144)=97200 J
W= 330750-97200=233550 J
233550J=235F
F=994N
 

berkeman

Mentor
55,901
5,958
Holy smokes. Your prof/TA is messing with you.

-1- right answer, wrong math. What does the phrase "gravity is a conservative force field" mean? Plus, what does the sun have to do with gravity on the Earth?

-3- The Earth has a speed of 2.97m/s with respect to what???
 
682
1
If the earth moves around in a circle, then there could not be any work done by the sun ! since the earth's kinetic energy and potential energy stays constant.
 

Doc Al

Mentor
44,815
1,078
1) I said none because the disatnce in the equation force times distance has to be a parallel distance to the force exerted on it, and its not in this case.
Good! The motion of the earth is perpendicular to the direction of the sun's force--so no work is done.

2)6.2m/3.2m/s=1.9375 sec
120N-Ff=m(0)
Ff=120N
W=Fd
W=120x6.2
W=744J
I have no idea why you calculated the time, but the rest is correct.

3)2(1.49 x10^11m)pi
d=2.97m
(3.56 x10^11 N)(2.97m)=1.06 x10^12 J
What happened here? Review your answer for question 1. (FYI: What's that speed of 2.97m/s supposed to be? The speed of the earth about the sun is about 10000 times greater.)

4)(1350kg)(9.8)(25m)
Ep'=330750 J
Ek2=(0.5)1350(144)=97200 J
W= 330750-97200=233550 J
233550J=235F
F=994N
Good.
 

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