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

In summary: So basically, in summary, 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 on a horizontal surface. The total work done on the cart is 101 joules. If the cart starts from rest, its final speed will be 4.49 meters per second. Assuming a coefficient of friction (µ) between the cart and surface, for µ to just bring the cart to a halt after 11 meters, the value of µ would be 0.259. The speed of the cart after it has gone
  • #1
r34racer01
63
0
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?

 
Physics news on Phys.org
  • #2
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)
 
  • #3
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?
 
  • #4
Never mind I got it, I was plugging in the wrong numbers.
 

What is work and how is it calculated?

Work is defined as the measure of the energy transferred when a force is applied to an object and it moves in the direction of the force. It is calculated by multiplying the magnitude of the force by the distance the object moves.

How is work related to friction?

Friction is a force that acts in the opposite direction of motion and can decrease the amount of work done on an object. In the case of a frictionless surface, no work is done on the object as there is no force acting against motion.

What is the significance of the 10kg cart in this scenario?

The mass of an object affects the amount of work done on it. In this scenario, a 10kg cart is being used, meaning that the force required to move it a certain distance will be greater than a lighter object, resulting in more work being done.

Why is the surface considered frictionless?

A frictionless surface is one that has no resistance to motion. This is often used in theoretical or ideal scenarios to simplify calculations and eliminate the effects of friction on an object.

How is the work done on the cart affected by the distance it moves?

The work done on an object is directly proportional to the distance it moves. In this scenario, the cart moves a distance of 11m, so the work done on it will be greater than if it had only moved a shorter distance.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
5K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
5K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top