Calculating Work, Force, and Power in Bob's Rock Throw

In summary, the question asks for the work, force, and power exerted by Bob as he throws a 620g rock with a speed of 30m/s, while moving his hand forward .8m. The work done on the rock can be found using the work-energy theorem, and the force and power can be calculated using the equations W = Fd and P = W/t.
  • #1
aligass2004
236
0

Homework Statement



Bob can throw a 620g rock with a speed of 30m/s. He moves his hand forward .8m while doing so.
a.) How much work does Bob do on the rock?
b.) How much force, assumed to be constant, does Bob apply to the rock?
c.) What is Bob's maximum power output as he throws the rock?

Homework Equations



W = Fd
P = W/t

The Attempt at a Solution



For part a, I tried using W = Fd. I assumed F would be mg, but I think that the accelerating has to be factored in there somewhere as well. I got 4.861, but it was wrong.
 
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  • #2
You're not given the acceleration of the rock, however you do know its mass and its final speed so to speak. Assuming the rock accelerates from rest, we can use the work-energy theorem to find the work done on the rock. So we have:

[tex] W_{TOT} = \Delta KE [/tex]

Does that help?
 
  • #3
I know that for K, I would substitute 1/2m(v^2), but I don't know how to find E.
 
  • #4
aligass2004 said:
I know that for K, I would substitute 1/2m(v^2), but I don't know how to find E.

I'm not sure what you mean by "E." You're asked to find work, force, and power. What are you stuck on exactly?
 
  • #5
Oh, oh...KE meant change in kinetic energy. I thought it meant kinetic energy times energy...I don't know why I thought that though.
 

1. How does Bob throwing a rock relate to work being done?

Throwing a rock involves applying a force to the rock to move it from one place to another. This act of applying force and causing displacement is known as work. Therefore, Bob throwing a rock involves work being done.

2. What factors affect the amount of work done when Bob throws a rock?

The amount of work done when Bob throws a rock is affected by the force applied to the rock, the distance the rock is thrown, and the direction of the force relative to the direction of the displacement.

3. How is the work done by Bob throwing a rock calculated?

The work done is calculated by multiplying the force applied by the distance the rock is thrown, and then taking into account the angle between the force and the displacement using the formula W = F * d * cos(theta).

4. Does the weight of the rock affect the work done when Bob throws it?

Yes, the weight of the rock does affect the work done because it is a factor in determining the force applied. The heavier the rock, the more force Bob has to apply to throw it, resulting in more work being done.

5. How does the work done by Bob throwing a rock change if the rock is thrown at an angle instead of straight ahead?

When the rock is thrown at an angle, the force is applied in a direction that is not directly in line with the displacement. This means that less work is done compared to throwing the rock straight ahead, as the force is not fully used to cause displacement. The amount of work done can be calculated by using the cosine of the angle between the force and displacement in the formula W = F * d * cos(theta).

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