Power

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Physics › Power

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To accelerate your car at a constant acceleration, the car’s engine must

Maintain a constant turning speed

Maintain a constant power output

Develop ever decreasing power

Develop ever increasing power

Explanation

Power is equal to the Work put into the system per unit time.

Work is equal to the force acting on the object multiplied by the displacement through which it acts.

Therefore power is directly related to the force applied.

Force is also directly related to the acceleration of an object. A constant force will create a constant acceleration.

Since power is directly related to the force applied, and the force must be constant to maintain a constant acceleration, the power must also therefore be constant.

To accelerate your car at a constant acceleration, the car’s engine must

Maintain a constant turning speed

Maintain a constant power output

Develop ever decreasing power

Develop ever increasing power

Explanation

Power is equal to the Work put into the system per unit time.

Work is equal to the force acting on the object multiplied by the displacement through which it acts.

Therefore power is directly related to the force applied.

Force is also directly related to the acceleration of an object. A constant force will create a constant acceleration.

Since power is directly related to the force applied, and the force must be constant to maintain a constant acceleration, the power must also therefore be constant.

A sports car accelerates from rest to in . What is the average power delivered by the engine?

Explanation

Power is equal to the work done divided by how much time to complete that work.

$P = \frac{W}{t}$

Work is equal to the change in kinetic energy of an object.

$W = \Delta KE$

$KE = \frac{1}{2}mv^2$

We can solve for the work by solving for the change in kinetic energy.

Since the initial velocity is , this can be dropped from the equation.

needs to be converted to

$\frac{90km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s} = 25m/s$

$W = \frac{1}{2}(1000kg)(25m/s)^2$

We can now plug this into the power equation.

$P = \frac{312500J}{6s}$

The average power is Watts.

The quantity is

The kinetic energy of the object

The potential energy of the object

The work done on the object by the force

The power supplied to object by the force

Explanation

Power is equal to the work divided by the time to complete the work.

$P = \frac{W}{t}$

Work is equal to the force times the displacement through which the object moved.

We can substitute this into our power equation to get

$P = \frac{Fd}{t}$

Velocity is equal to the distance over the time.

$v = \frac{d}{t}$

Therefore power could be written as

A sports car accelerates from rest to in . What is the average power delivered by the engine?

Explanation

Power is equal to the work done divided by how much time to complete that work.

$P = \frac{W}{t}$

Work is equal to the change in kinetic energy of an object.

$W = \Delta KE$

$KE = \frac{1}{2}mv^2$

We can solve for the work by solving for the change in kinetic energy.

Since the initial velocity is , this can be dropped from the equation.

needs to be converted to

$\frac{90km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s} = 25m/s$

$W = \frac{1}{2}(1000kg)(25m/s)^2$

We can now plug this into the power equation.

$P = \frac{312500J}{6s}$

The average power is Watts.

The quantity is

The kinetic energy of the object

The potential energy of the object

The work done on the object by the force

The power supplied to object by the force

Explanation

Power is equal to the work divided by the time to complete the work.

$P = \frac{W}{t}$

Work is equal to the force times the displacement through which the object moved.

We can substitute this into our power equation to get

$P = \frac{Fd}{t}$

Velocity is equal to the distance over the time.

$v = \frac{d}{t}$

Therefore power could be written as

Some electric power companies use water to store energy. Water is pumped from a low reservoir to a high reservoir. To store the energy produced in hour by a electric power plant, how many cubic meters of water will have to be pumped from the lower to the upper reservoir?

Assume the upper reservoir is an average of above the lower. Water has a mass of $1.00*10^{3}kg$ for every $1.0m^{3}$ .

$4.93*10^{5}m^{3}$

Explanation

First, we need to calculate the amount of energy produced by the power plant.

$P = \frac{energy}{time}$

$180*10^6 Watts = \frac{energy}{1hr} * \frac{1hr}{3600s}$

We know that this energy is stored in the form of gravitational potential energy.

$\frac{6.48*10^8 Joules}{(9.8m/s^2)(380m)} = m$

If there is for every of water, we can divide this number by to determine the number of of water is pumped from the lower to the upper reservoir.

$\frac{1.74*10^8 kg}{1000kg/m^3} = 1.74*10^5m^3$

Assume the upper reservoir is an average of above the lower. Water has a mass of $1.00*10^{3}kg$ for every $1.0m^{3}$ .

$4.93*10^{5}m^{3}$

Explanation

First, we need to calculate the amount of energy produced by the power plant.

$P = \frac{energy}{time}$

$180*10^6 Watts = \frac{energy}{1hr} * \frac{1hr}{3600s}$

We know that this energy is stored in the form of gravitational potential energy.

$\frac{6.48*10^8 Joules}{(9.8m/s^2)(380m)} = m$

If there is for every of water, we can divide this number by to determine the number of of water is pumped from the lower to the upper reservoir.

$\frac{1.74*10^8 kg}{1000kg/m^3} = 1.74*10^5m^3$

Of the following, which is not a unit of power?

watt/second

newton meter /second

joule/second

Watt

Explanation

The unit for power is the Watt. A watt is a measure of the Joules per second that an object uses. Joules can also be written as Newton Meters. Therefore a watt could also be considered a Newton Meter/Second. The incorrect answer is the watt/second since watt is the base unit of power on its own.

Of the following, which is not a unit of power?

watt/second

newton meter /second

joule/second

Watt