PHYSICS work/power/energy question

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In summary, we discussed several concepts related to work and power, including the relationship between net force and work, work being done on particles at rest, and the effect of perpendicular forces on work. We also examined how to find power at any given time and the work done by a force over a period of time. Finally, we used these concepts to determine the power required to operate a ski lift.
  • #1
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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  • #2
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 ([itex]W=P\Delta V[/itex]).
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 [itex]W = \int \vec{F}\cdot \vec{s} = \int F\cdot \vec{v} dt = \int Fvdt cos\theta[/itex] if [itex]\theta[/itex] 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 [itex]P = dW/dt = Fds/dt = Fv = mav = m\ddot x \dot x[/itex]
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
 
  • #3
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 ([itex]W=P\Delta V[/itex]).
See example above.
True. Since [itex]W = \int \vec{F}\cdot \vec{s} = \int F\cdot \vec{v} dt = \int Fvdt cos\theta[/itex] if [itex]\theta[/itex] is always 90 degrees, then W = 0.

Use [itex]P = dW/dt = Fds/dt = Fv = mav = m\ddot x \dot x[/itex]
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...
 
  • #4
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
 

1. What is the definition of work in physics?

In physics, work is the measure of energy transfer that occurs when a force is applied to an object and causes it to move in the direction of the force.

2. How is work calculated in physics?

The formula for calculating work in physics is W = F * d, where W is work, F is the force applied, and d is the distance moved in the direction of the force.

3. What is the relationship between work and energy in physics?

In physics, work and energy are closely related. Work is the transfer of energy, and energy is the ability to do work. This means that any work done on an object will result in a change in its energy.

4. How is power defined in physics?

Power in physics is the rate at which work is done or energy is transferred. It is calculated by dividing the amount of work done by the time it takes to do the work, or P = W/t.

5. What is the difference between potential energy and kinetic energy?

Potential energy is the energy an object has due to its position or configuration, while kinetic energy is the energy an object has due to its motion. Potential energy can be converted into kinetic energy and vice versa.

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