Work Done by a Constant Force (Q6 page 188 of Cutnell and Johnson)

In summary, the formula for calculating work (W = F cos Θ s) was used to determine the amount of work done on the toboggan by the tension force of 94.0 N pulling it for a distance of 35.0 m at an angle of 25.0°. The final answer is 2980 J.
  • #1
gcombina
157
3
6.A person pulls a toboggan for a distance of 35.0 m along the snow with a rope directed 25.0° above the snow. The tension in the rope is 94.0 N.

a) How much work is done on the toboggan by the tension force? Ok, so W = (F cos Θ) s
= (94.0 cos 25°) 25 m
= WRONG ANSWER

I know the answer is 94.0 times distance but that would equal WORK right (fs)

why are they doing (94.0 N)(35.0 M) = Force?

**this is really frustrating , why are they including distance as force?
 
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  • #2
So what does it say in terms of a "right" answer or solution?

Why does your solution have 94.0 twice?
 
  • #3
gcombina said:
6.A person pulls a toboggan for a distance of 35.0 m along the snow with a rope directed 25.0° above the snow. The tension in the rope is 94.0 N.

a) How much work is done on the toboggan by the tension force? Ok, so W = (F cos Θ) s
= (94.0 cos 25°) 94.0 N
Hi gcombina. What does that "s" in your formula stand for? What units would it have?

Also, what are the units of that "W"?
 
  • #4
this is the "right" answer

W = (94.0 N) (35.0 m) cos 25.0 = 2980 J
 
  • #5
olivermsun said:
So what does it say in terms of a "right" answer or solution?

Why does your solution have 94.0 twice?


it was my mistake, i corrected it already.
 
  • #6
NascentOxygen said:
Hi gcombina. What does that "s" in your formula stand for? What units would it have?

Also, what are the units of that "W"?
basically, this is the right answer => (94.0 N) (35.0 m) cos 25° =2980 J

this is formula (F cos θ)s I always use, but now they are using (Fs cos θ), where is this formula comes from?

I though W = (F cos θ)s and NOT (Fs cos θ)
 
  • #7
Uhmmmm ... It is the same. Multiplication is commutative so it does not matter what order you multiply things in.
 
  • #8
gcombina said:
basically, this is the right answer => (94.0 N) (35.0 m) cos 25° =2980 J

this is formula (F cos θ)s I always use, but now they are using (Fs cos θ), where is this formula comes from?

I though W = (F cos θ)s and NOT (Fs cos θ)
Do these give the same answer?
 
  • #9
NascentOxygen said:
Do these give the same answer?

no, they don't give the same answer

Fs cos θ => (94.0 N) (35.0 m) cos 25° =2980 J

AND

F cos θ => (94.0 N) cos 25° = 85.19 J
 
  • #10
gcombina said:
F cos θ => (94.0 N) cos 25° = 85.19 J

Check your units. You wrote Joules but it is actually Newtons.

I think you meant "(F cos θ)s"? In which case, you do get the same answer.
 
  • #11
gcombina said:
Ok, so W = (F cos Θ) s
= (94.0 cos 25°) 25 m
The distance is 35m, not 25m.
 
  • #12
There is something very fundamental that you are misunderstanding, but I'm having difficulty identifying what.

gcombina said:
why are they doing (94.0 N)(35.0 M) = Force?

**this is really frustrating , why are they including distance as force?

They aren't. It's difficult to see why you might think they are.

F s cosθ = Force x distance x cosθ
 
  • #13
Starting from the beginning, step by step:

W = (F cos Θ) s

The distance is s = 35.0 m
The force in the direction that the distance is measured over is F cos Θ = (94.0 N)(cos 25°)

W = (F cos Θ) s = (94.0 N)(cos 25°)(35.0 m) = 2980 N m = 2980 J.
 

1. What is work done by a constant force?

Work done by a constant force is the product of the magnitude of the force and the displacement of the object in the direction of the force. It is a measure of the energy required to move an object over a certain distance.

2. How is work calculated for a constant force?

The formula for calculating work done by a constant force is W = Fd cosθ, where W is work, F is the magnitude of the force, d is the displacement, and θ is the angle between the force and displacement vectors.

3. What is the unit of measurement for work?

The unit of measurement for work is the joule (J), which is equivalent to kg*m^2/s^2.

4. Can work done by a constant force be negative?

Yes, work done by a constant force can be negative if the force and displacement vectors are in opposite directions. This means that the force is acting against the motion of the object.

5. How does the angle between the force and displacement vectors affect the work done?

The angle between the force and displacement vectors affects the work done by changing the component of the force that is in the direction of the displacement. When the angle is 0 degrees (cosθ = 1), the force is fully in the direction of the displacement and does the most work. When the angle is 90 degrees (cosθ = 0), the force is perpendicular to the displacement and does no work.

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