Work Done on 10kg Cart on Frictionless Surface: 11m

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Homework Help Overview

The discussion revolves around a physics problem involving a 10 kg cart being pulled and pushed on a frictionless surface, with various forces applied over a distance of 11 meters. The participants are exploring the net work done on the cart and its resulting speed, as well as the effects of friction when introduced into the scenario.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of total work done on the cart, questioning the application of work-energy principles and the effects of different forces acting on the cart.
  • There are attempts to derive the coefficient of friction that would bring the cart to a halt after the specified distance, with some participants expressing confusion over their calculations.

Discussion Status

Some participants have provided insights into the calculations, particularly regarding the direction of forces and the total work done. There is ongoing exploration of the correct approach to finding the coefficient of friction, with some participants indicating they have resolved their earlier confusion.

Contextual Notes

Participants are working under the assumption of a frictionless surface initially, then introducing friction into the problem. There are indications of miscalculations and misunderstandings regarding the application of formulas related to work and energy.

r34racer01
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A 10 kg cart is pulled by a force of 14 Newtons for a distance of 6 meters, then by a force of 6 Newtons for 4 meters, then it is pushed (slowed down) by a force of 6 Newton for 1 meters. All this occurs on a horizontal surface.

a) Initially assuming the surface to be frictionless, what net work is done on the cart as it moved the 11 meters?
b) If the cart starts from rest, what is its speed after it has moved the 11 m?

Now assume a coefficient of friction µ between cart and surface during the entire distance.

c) For what value of μ does the cart just come to a halt after 11 m?
d) For the value of µ computed in part (c), what is the speed of the cart after it has gone 4 m?


W = ΔKE
W = Fd cosθ
Ff = μN
ΣF = ma
KE = (1/2)mv^2


So for a.) I got 101 joules by dumb luck. I thought I could do Wtot = (14*6)+(6*4)+(6*1) = 114, but that didn't work and I just entered 101 at random. Can someone tell me how to actually calculate the total work?
For b.) I knew Wtot = (1/2)mv^2, so 101=5v^2 => Vf= 4.49

c.) is where I'm really stuck. I thought I could do Wtot = Fd = μ*m*g*d => μ = 0.094
but that's wrong. Can someone tell me how to get this part please?

 
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Hi r34racer01! :smile:
r34racer01 said:
A 10 kg cart is pulled by a force of 14 Newtons for a distance of 6 meters, then by a force of 6 Newtons for 4 meters, then it is pushed (slowed down) by a force of 6 Newton for 1 meters. All this occurs on a horizontal surface.

So for a.) I got 101 joules by dumb luck. I thought I could do Wtot = (14*6)+(6*4)+(6*1) = 114, but that didn't work and I just entered 101 at random. Can someone tell me how to actually calculate the total work?

The third force is backwards :wink:

(but I still don't make it 101)
 
tiny-tim said:
Hi r34racer01! :smile:


The third force is backwards :wink:

(but I still don't make it 101)

Yeah apparently it is actually W = 102 for pt a.), weird right. Anyway I'm still having trouble w/ pt c.). It should be 102 = μmgd = 0.259 but its telling me that's wrong. Help anyone?
 
Never mind I got it, I was plugging in the wrong numbers.
 

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