Power and conservation of energy

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Homework Help Overview

The discussion revolves around two physics problems related to power and conservation of energy. The first problem involves a monkey and sled moving up an inclined track, requiring an understanding of energy conservation and friction. The second problem pertains to calculating the running speed of a person based on energy dissipation and power output.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of conservation of energy principles, questioning how to incorporate friction into their calculations. There are attempts to relate kinetic and potential energy, as well as discussions on how to derive force from power and energy dissipation.

Discussion Status

Some participants have provided hints and guidance on how to approach the problems, particularly emphasizing the need to account for friction in the first problem. There is a recognition of the complexity introduced by angles and the need for further clarification on resolving forces.

Contextual Notes

Participants express difficulty with the angle components in the first problem and the way energy dissipation is presented in the second problem. There is an acknowledgment of the need for clearer definitions and setups to facilitate understanding.

gosabres
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Hi! I'm having a hard time with these two problems. Any help would be greatly appreciated.

1. in a circus performance, a monkey is strapped to a sled and both are given an initial speed of 2.0 m/s up an 18 degree inclined track. The combined mass of the monkey and sled is 20 kg, and the coefficient of kinetic friction between the sled and incline is 0.20. How far up the incline do the monkey and sled move?


2. While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 70 kg person develops a power of 73 W during a race, how fast is the person running? Assume a running step is 1.5 m long.
 
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https://www.physicsforums.com/showthread.php?t=94379

Post what you have tried.
 
Hi! Sorry I didn't show work. I just seriously don't know how to start either one of these.
For the monkey problem, i think i have to somehow use the conservation of energy equation (1/2 mvi^2 + mgyi = 1/2 mvf^2 + mgyf) to figure out the distance. i just don't even know what steps to take.
Same with the power problem P = Fv but I'm just really lost. I'll keep trying. Any hints to help me get started? If not, thanks anyway.
 
gosabres said:
Hi! Sorry I didn't show work. I just seriously don't know how to start either one of these.
For the monkey problem, i think i have to somehow use the conservation of energy equation (1/2 mvi^2 + mgyi = 1/2 mvf^2 + mgyf) to figure out the distance. i just don't even know what steps to take.
Same with the power problem P = Fv but I'm just really lost. I'll keep trying. Any hints to help me get started? If not, thanks anyway.
Energy is a good approach to the first problem, but you have to expand conservation to include the work done by the frictional force. The mechanical energy (KE + PE) at the top will be less than the mechanical energy at the bottom. The difference between them is the work done by friction. How is work related to force? Can you find the force of friction?

The second problem has been obscured by giving you the energy dissipation per step instead of per meter, and by telling you information about how much work is done to move a certain distance instead of the force applied. Use the work per distance information to find the force. Then use the power and force to find velocity.
 
Hi! So... i was able to figure out the power problem pretty easily using your suggestions(Thanks!). I came up with a final answer of 2.6 m/s.

But... the monkey problem is tripping me up big time. I've tried a lot of different things. This is my latest attempt:

- ukmgd = 1/2 mvf^2 - 1/2 mvi^2
- (.20) (20) (9.8) d = 0 - 1/2 (20) (2.0 cos18)^2
d = 0.92 m

i don't know if I'm still way off base or if I'm just messing up the angle part?
probably both... but I've noticed i have the most trouble with problems that have angles.

Thanks for any help,
Kelly
 
gosabres said:
Hi! So... i was able to figure out the power problem pretty easily using your suggestions(Thanks!). I came up with a final answer of 2.6 m/s.

But... the monkey problem is tripping me up big time. I've tried a lot of different things. This is my latest attempt:

- ukmgd = 1/2 mvf^2 - 1/2 mvi^2
- (.20) (20) (9.8) d = 0 - 1/2 (20) (2.0 cos18)^2
d = 0.92 m

i don't know if I'm still way off base or if I'm just messing up the angle part?
probably both... but I've noticed i have the most trouble with problems that have angles.

Thanks for any help,
Kelly
Again energy is a good approach. The work done by friction will reduce the mechanical energy. You need to resolve the weight into components to find the normal force between the track and the sled/monkey combo. From the normal force and coefficient of friction, find the friction force. Relate the distance moved up the track to the change in elevation. The initial energy (kinetic) equals the final energy (PE) plus the work done by the frictional force.
 
The difficult part about this monkey prob is that energy is not conserved so u need to calculate the energy loss due to friction.

Therefore, the equation in this case would be..
Kinetic energy=Potential energy+energy loss due to friction.

To find kinetic friction, all u need is to have the frictional force multiply by the overall distance traveled in the direction of motion of the sled. Perhaps to make this a little simpler, you can try drawing out the free body diagram on the sled. Try breaking up the weight into its horizontal and vertical forces. You will be able to find that the horizontal component of the weight will actually be ur frictional force! Since u do not know the distance travelled, maybe leave it in height/sin 18. Subbed the values of kinetic friction and distance into finding ur work done and solve for h. Once you solve for h, use pythagoras theorem to solve for the distance traveled up the incline.
 
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