Ramp and friction problem

[chaser0]

A 2-kg cart starts on a 2m high table before it slides down a 30 degree ramp with coefficient of friction 0.20 and then comes to a complete rest 1.31m from the ramp on a horizontal surface. How much work was done to stop the car?

Its a discrete problem, no diagram.


Im trying to solve this, but the friction coefficient is bothering me.

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[chiddler]

A 2-kg cart starts on a 2m high table before it slides down a 30 degree ramp with coefficient of friction 0.20 and then comes to a complete rest 1.31m from the ramp on a horizontal surface. How much work was done to stop the car?

Its a discrete problem, no diagram.


Im trying to solve this, but the friction coefficient is bothering me.

friction is the only thing stopping the cart. so you can solve this a few ways:

1. mgh = energy at the top of the ramp. all this energy is converted to thermal via friction. so it's (2)(10)(2) = 40 J of work done by friction.

2. or you can do W = F*d. What is force? The only force moving it is gravity so it's mgsin(theta)=2(10)*0.5 = 10 N. What is distance? Triangle trig: sin(30) = 2 / d; d = 4.

F*d = 10 * 4 = 40 J.

yeah!

 

[Buttafuoco]

I always like to use KE = PE for these types of problems. It's so much easier to find work, velocity, anything starting with its potential energy than effing around with trig values.

EDIT obviously this isn't really a case where you use KE at all, but still i just mean using the potential energy to do the problem.

 

[chaser0]

I understand the technique where you completely circumvent the coefficient of friction and just do PE=W.
The easier way.


But I was wondering about this path

2. or you can do W = F*d. What is force? The only force moving it is gravity so it's mgsin(theta)=2(10)*0.5 = 10 N. What is distance? Triangle trig: sin(30) = 2 / d; d = 4.

If a box moves from point A to point B, and there is a large coefficient of friction, is there not MORE work exhibited on the box compared to if it was frictionless?

 

[Buttafuoco]

if there's a larger coefficient of friction, it will stop sooner. It can't dissipate more energy than it had, and all its energy is due to gravity.

 

[chaser0]

if there's a larger coefficient of friction, it will stop sooner. It can't dissipate more energy than it had, and all its energy is due to gravity.

awesome thanks

 

[chiddler]

I understand the technique where you completely circumvent the coefficient of friction and just do PE=W.
The easier way.


But I was wondering about this path



If a box moves from point A to point B, and there is a large coefficient of friction, is there not MORE work exhibited on the box compared to if it was frictionless?

wait! I just realized i'm wrong because I forgot to account for the extra 1.31 meters traveled after the ramp.

yeah i'm not sure. can anybody please correct my work?

 

[syoung]

you're right. all the potential energy at the top is what it starts with, and friction eats up all kinetic energy it builds up as it goes down the ramp and forward. it doesnt matter how far the car travels past the ramp, because it cannot gain more energy than what it started with (40J). the kinetic friction just tells you how fast or slow it will "slow" down,

 

[chiddler]

yes from W=mgh standpoint, that works well. However, from the W=F*d calculations I did, that's completely wrong because the d i used was 4 where it should be 5.31.

Oh hey! it's because I used the wrong force! i forgot to subtract friction force from gravitational force....

OP i hope you caught this stupid mistake.