# Finding the power and work needed

• foozeball2468
In summary: The direction of the push force will also change as the floor slopes upward.(solving for t I get 11m/.62m/s=17.74 s) so W=448x17.74 which is 7.96 x10^3 J -both of these answers are smaller so I do not believe this is trueIn summary, the original problem is about determining the power needed to push a 95 kg chest at 0.62 m/s along a horizontal floor with a coefficient of friction of 0.78. The answer is 450W. For the second question, where the floor slopes upward at an angle of 5 degrees and the chest is pushed parallel to the floor, two different approaches were
foozeball2468
This is the orginal problem...a.)How much power is needed to push a 95 kg chest at 0.62 m/s along a horizontal floor where the coefficient of friction is 0.78.(ans. 450W)... b.) How much work is done in pushing the chest 11 m? (ans. 7.99x10^3 J)

New question: Answer the previous question when the floor slopes upward at an angle of 5 degrees, and you push parallel to the floor.

Fa=(coefficient of friction)(mg)
P=(F) (v)
W=(F) (delta x)

My attempt is W=(95kg)(9.81m/s^2)(11m)(sin5 +cos5(.78))= 8.81 x10^3 J so solving for power I would use Work/17.74= 496 Watts
.
A different route is P=(.78)(95kg)(9.8m/s^2)(cos5)(.62)=448 W
(solving for t I get 11m/.62m/s=17.74 s) so W=448x17.74 which is 7.96 x10^3 J
-both of these answers are smaller so I do not believe this is true
I am unsure if either of these is right and if both are wrong where have I gone wrong! thank you

foozeball2468 said:
This is the orginal problem...a.)How much power is needed to push a 95 kg chest at 0.62 m/s along a horizontal floor where the coefficient of friction is 0.78.(ans. 450W)... b.) How much work is done in pushing the chest 11 m? (ans. 7.99x10^3 J)

New question: Answer the previous question when the floor slopes upward at an angle of 5 degrees, and you push parallel to the floor.

Fa=(coefficient of friction)(mg)
P=(F) (v)
W=(F) (delta x)

My attempt is W=(95kg)(9.81m/s^2)(11m)(sin5 +cos5(.78))= 8.81 x10^3 J so solving for power I would use Work/17.74= 496 Watts
this looks good
.
A different route is P=(.78)(95kg)(9.8m/s^2)(cos5)(.62)=448 W
what happened to the sin5 piece... ?? The push force and power required must overcome both friction and gravity.

## What is power?

Power is the rate at which work is done or energy is transferred. It is measured in watts (W) and is calculated by dividing the amount of work done by the time it takes to do that work.

## What is work?

Work is the transfer of energy to an object by applying a force over a distance. It is measured in joules (J) and is calculated by multiplying the force applied by the distance over which it is applied.

## How do you calculate power?

Power is calculated by dividing the amount of work done by the time it takes to do that work. It can also be calculated by multiplying force by velocity or by using the formula P=VI where V is voltage and I is current.

## What is the unit of measurement for power?

The unit of measurement for power is watts (W). Other commonly used units include horsepower (hp) and kilowatts (kW).

## What factors affect the amount of power and work needed?

The amount of power and work needed can be affected by factors such as the force applied, the distance over which the force is applied, the time it takes to do the work, and the efficiency of the system performing the work.

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