Energy and Work Concept. Confusing. Help please

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Energy and Work Concept. Confusing. Help please :)

1. Homework Statement
A train composed of flat cars, travels with s constant speed V along a straight
track. A crate of mass m is attached to a rope and is slowly lowered from a
bridge when the train passes under it. At the instant that the crate touches the
train, the rope is released. the create then slides some distance along the train
and after some time T, comes to rest relative to it. In answering the following
questions, describe all quantities relative to the reference frame of the bridge,
a) Is the work done on the crate, by the force due the train, positive, negative
or zero?
b) Is the work done on the train, by the force due to the crate, positive,
negative or zero?
c) Is the magnitude of the work done on the crate, by the force due to the
train, larger than, smaller than, or equal to the magnitude of the work done
on the train by the force due to the crate? Explain your answer.
d) What is the distance, which the crate slides, relative to the train before
coming to rest relative to it?
e) During the time T, what is the work on done on crate by the train?
f) During the time T, what is the work done on the train by the crate?
g) What is the coefficient of kinetic friction between the crate and the train

Homework Equations


Nothing else than energy and work equations. No momentum is allowed yet unfortunately.

The Attempt at a Solution


I got really confused while deciding the direction of the forces, whether both forces are friction forces or not. I need help as this is a concept question and will improve my thinking. Thanks :smile::confused:
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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Welcome to PF!
The friction force accelerates the crate (in the direction of the train), doing work on it, giving it kinetic energy. An equal force is exerted backward on the train.
 
  • #3
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Hi and thank you! :)

So that means;
a) Work done on crate is positive

b) Work done on train is negative

c) Magnitudes of the works are equal since magnitudes of forces are equal.

d) So the velocity of the train decreases as the crate deccelerates it, doesnt it? If Vi is the initial speed, Vf is the final speed of the train, f is the friction force,m is the mass of crate and M is the mass of train;
d=[1/2*M*Vi^2-1/2*(M+m)*Vf^2]/f

e) I need help with this part.

f) f*d. Is it this simple?

g) From f*d=1/2*M*Vi^2-1/2*(M+m)*Vf^2
f=[1/2*M*Vi^2-1/2*(M+m)*Vf^2]/d
f=m*g*uk
uk=[1/2*M*Vi^2-1/2*(M+m)*Vf^2]/(d*m*g)
 
  • #4
Delphi51
Homework Helper
3,407
10


c) I think this is more complicated. It can't be just energy gained by crate = energy lost by train, because the friction implies loss of energy to heat.

d)The question says the train moves at constant speed; it isn't clear whether it slows when the crate lands on it - perhaps the engine makes up the loss of energy to heat. Anyway, the slowing would be negligible. Your answer to d) should surely depend on T and I don't think it should depend on f, since f isn't given. Won't it just be d = (average v)*T ?

e) "work done on crate". I would say just its gain in kinetic energy, going from speed 0 to v.
f) I think you have to add the energy lost to heat through the friction. This is not entirely clear to me!
 
  • #5
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Hmm well,
c) energy gained by crate < energy lost by train because train lost more energy in the means of heat due to friction force.

d) what about Vavg ?

e) as the speed change is neglected V is equal to Vi,thus the work done on crate is;
1/2*m*Vi^2

f) but can friction be counted as work done by crate ?
 
  • #6
Delphi51
Homework Helper
3,407
10


The friction force will be constant, so constant acceleration. Use constant acceleration formulas to find the average velocity of the crate.

I think "work done on the train by the crate" would include friction. In fact, the only force on the train is the friction force.
 

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