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

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A 10 kg cart is pulled and pushed on a frictionless surface, with net work calculated to be 102 joules after moving 11 meters. The speed of the cart after this distance, starting from rest, is determined to be approximately 4.49 m/s. When considering friction, the coefficient of friction needed to bring the cart to a halt after 11 meters is calculated, but initial attempts yield incorrect values. The correct approach involves using the work-energy principle and ensuring accurate values are plugged into the equations. Ultimately, the discussion highlights the importance of careful calculations in physics problems involving forces and motion.
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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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