Work Done Dragging 50kg Crate 8m: Find Force, Angle & Energy

  • Thread starter lynchdemartin
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In summary, a 50kg crate is dragged at a constant velocity of 1.5m/s for a distance of 8m. If the force of friction on the crate is 225N, the person does 1690J of work.
  • #1
lynchdemartin
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A 50.0 kg crate is dragged at a constant velocity of 1.5 m/s for a distance of 8.0 m. If the force of friction on the crate is 225 N, how much work was done by the person to drag the crate over the 8.0 m distance if the rope used to drag the crate was directed at 20degrees angle above the horizontal?

my work:
mass 50kg
distance 8m
angle 20 degrees
Force of Friction 225N
since we have a constant velocity Fnet = 0
Fnet = T - Ff

Solve for T
0 = T - 225N
T = 225N

therefore T x distance moved by the rope x cosx = work done
so I got 225N x 8m x cos20 = 1690 J

However the answer sheet says it's 1800 J...
 
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  • #2
lynchdemartin said:
A 50.0 kg crate is dragged at a constant velocity of 1.5 m/s for a distance of 8.0 m. If the force of friction on the crate is 225 N, how much work was done by the person to drag the crate over the 8.0 m distance if the rope used to drag the crate was directed at 20degrees angle above the horizontal?

my work:
mass 50kg
distance 8m
angle 20 degrees
Force of Friction 225N
since we have a constant velocity Fnet = 0
Fnet = T - Ff

Solve for T
0 = T - 225N
T = 225N

therefore T x distance moved by the rope x cosx = work done
so I got 225N x 8m x cos20 = 1690 J

However the answer sheet says it's 1800 J...

The tension T is greater than 225N, since the horizontal force (the cos component of T) is 225N to oppose the friction force.
 
  • #3
berkeman said:
The tension T is greater than 225N, since the horizontal force (the cos component of T) is 225N to oppose the friction force.

That's only if the object is "not moving at constant velocity". In this question it is moving at constant velocity so Fnet must equal 0.
 
  • #4
lynchdemartin said:
That's only if the object is "not moving at constant velocity". In this question it is moving at constant velocity so Fnet must equal 0.

Correct. The Fnet in the horizontal direction has to equal zero (225 -225 = 0).
 
  • #5
I still get 1690J for work done. Driving me nuts...
 
  • #6
lynchdemartin said:
A 50.0 kg crate is dragged at a constant velocity of 1.5 m/s for a distance of 8.0 m. If the force of friction on the crate is 225 N, how much work was done by the person to drag the crate over the 8.0 m distance if the rope used to drag the crate was directed at 20degrees angle above the horizontal?

my work:
mass 50kg
distance 8m
angle 20 degrees
Force of Friction 225N
since we have a constant velocity Fnet = 0
Fnet = T - Ff

Solve for T
0 = T - 225N
T = 225N

therefore T x distance moved by the rope x cosx = work done
so I got 225N x 8m x cos20 = 1690 J

However the answer sheet says it's 1800 J...

You may confuse last part.
net F=0
Tcos20 = f =225 N
W=(Tcos20)*8=1800 J
 
  • #7
umm this might seem stupid but I think the question is asking what is the work done by the "person" not the "rope" so then it's T 225 N x 8m = 1800J ?
 
  • #8
lynchdemartin said:
umm this might seem stupid but I think the question is asking what is the work done by the "person" not the "rope" so then it's T 225 N x 8m = 1800J ?

The person is doing the work, but you've got the correct answer. The horizontal component of T is 225N, to oppose the frictional force. The actual tension T is greater than 225N, since there is also a vertical component to the Tension.
 
  • #9
berkeman said:
The person is doing the work, but you've got the correct answer. The horizontal component of T is 225N, to oppose the frictional force. The actual tension T is greater than 225N, since there is also a vertical component to the Tension.

thanks I can finally rest now..
 

1. What is work?

Work is defined as the force applied over a distance. In other words, it is the energy transferred when a force is used to move an object.

2. How do you calculate work?

Work is calculated by multiplying the force applied by the distance moved in the direction of the force. In this scenario, the work done to drag the 50kg crate 8m would be calculated by multiplying the force used by 8m.

3. What is the force required to drag a 50kg crate 8m?

The force required to drag a 50kg crate 8m would depend on the surface the crate is being dragged on and the friction present. To calculate the force, you would need to divide the work done by the distance moved.

4. How can the angle of the force be determined?

The angle of the force can be determined by using trigonometry. The angle would be the inverse tangent of the vertical distance (height) divided by the horizontal distance (length). In this scenario, the angle would be the inverse tangent of the height (0) divided by the length (8m), which would result in an angle of 0 degrees.

5. How much energy is required to drag a 50kg crate 8m?

To determine the amount of energy required to drag a 50kg crate 8m, you would need to know the force used and the distance moved. The energy would be equal to the work done, which is calculated by multiplying the force by the distance. In this scenario, the energy required would be the same as the work done to drag the crate 8m.

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