# Dynamics Question

#### lynchdemartin

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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#### berkeman

Mentor
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.

#### lynchdemartin

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.

#### berkeman

Mentor
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).

#### lynchdemartin

I still get 1690J for work done. Driving me nuts...

#### inky

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

#### lynchdemartin

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 ?

#### berkeman

Mentor
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.

#### lynchdemartin

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..

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