Several Energy Related Questions

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The discussion addresses several energy-related physics problems, focusing on work done by forces and energy transformations. It concludes that gravity does no work on Earth as its motion is perpendicular to the gravitational force. The calculations for a crate being pushed reveal that 744J of work is done, while the work-energy principle is applied to a car rolling down a hill, resulting in a friction force of 994N. Participants emphasize the importance of understanding the relationship between force, distance, and energy conservation in circular motion. The conversation highlights the need for clarity in calculations and concepts related to energy and forces.
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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?


Homework Equations


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

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
 
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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?
 
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.
 
Fusilli_Jerry89 said:
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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