PHYSICS work/power/energy question

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The discussion revolves around physics concepts related to work, power, and energy. Participants confirm that only the net force can do work, that no work is done on a particle at rest, and that a force perpendicular to velocity does not perform work. Calculations for power and work done by a particle moving along a defined path are discussed, with suggestions to use integration for work. A participant calculates the average power required for a ski lift but initially miscalculates the total work done, which is clarified to be 165 kW for 100 passengers.
r3dxP
8. For each statement, write true or false.
A)only the net force acting on an object can do work.
i think its TRUE.
B)no work is done on a particle that remains at rest.
i think its TRUE.
C)a force that is always perpendicular to the velocity of a particle never does work on the particle.
i think its TRUE.

^ please correct me if I am wrong.. ^ thx.

6. A particle of mass m moves along the x axis. Its position varies with time according to x=2t^3 + 4t^2 where x is in meters and t is in seconds.
B)find the power delivered to the particle at any time t.
?
C)the work done by the force from t=0 to t=11
im guessing i have to integrate the equation 2t^3 + 4t^2 with 11 on top right of the integration sign and 0 on the bottom of the integration sign.
any comment will be hhelpful. thanks in advance!

10. If a ski lift raises 100 passengers averaging 660N in weight to a height of 150m in 60s, at constant speed, what average power is required of the force making the lift?
I got P = W/t = mgh/t = 660N*150m/60s = 1650W = 1.6kW
did i do this right?
 
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r3dxP said:
8. For each statement, write true or false.
A)only the net force acting on an object can do work.
i think its TRUE.
Think about two equal and opposite forces compressing a volume of gas. The net force is 0. But there is work done (W=P\Delta V).
B)no work is done on a particle that remains at rest.
i think its TRUE.
See example above.
C)a force that is always perpendicular to the velocity of a particle never does work on the particle.
i think its TRUE.
True. Since W = \int \vec{F}\cdot \vec{s} = \int F\cdot \vec{v} dt = \int Fvdt cos\theta if \theta is always 90 degrees, then W = 0.

6. A particle of mass m moves along the x axis. Its position varies with time according to x=2t^3 + 4t^2 where x is in meters and t is in seconds.
B)find the power delivered to the particle at any time t.
Use P = dW/dt = Fds/dt = Fv = mav = m\ddot x \dot x
C)the work done by the force from t=0 to t=11
im guessing i have to integrate the equation 2t^3 + 4t^2 with 11 on top right of the integration sign and 0 on the bottom of the integration sign.
See suggestion in B)

10. If a ski lift raises 100 passengers averaging 660N in weight to a height of 150m in 60s, at constant speed, what average power is required of the force making the lift?
I got P = W/t = mgh/t = 660N*150m/60s = 1650W = 1.6kW
did i do this right?
The work done in raising 100 passengers is W=mgh where mg = 100*660N and h = 150 m. So the correct answer is 165000 W. or 165 kW.

AM
 
Andrew Mason said:
Think about two equal and opposite forces compressing a volume of gas. The net force is 0. But there is work done (W=P\Delta V).
See example above.
True. Since W = \int \vec{F}\cdot \vec{s} = \int F\cdot \vec{v} dt = \int Fvdt cos\theta if \theta is always 90 degrees, then W = 0.

Use P = dW/dt = Fds/dt = Fv = mav = m\ddot x \dot x
See suggestion in B)

The work done in raising 100 passengers is W=mgh where mg = 100*660N and h = 150 m. So the correct answer is 165000 W. or 165 kW.

AM

um.. mg = 660 m^2/s^2.. then multiply this by h=150m. I don't understand how you could put 100 as its mass...
 
r3dxP said:
um.. mg = 660 m^2/s^2.. then multiply this by h=150m. I don't understand how you could put 100 as its mass...
660 N is the weight of one person (average). There are 100 people. So the total weight is 100*660.

AM
 
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