# Solving a Power Problem: Work & Inst. Power

• AnkhUNC
In summary, we have an object with a mass of 2.9 kg that accelerates horizontally and uniformly from rest to a speed of 14 m/s in 4.7 seconds. In this time interval, the work done on the object by the accelerating force is 284.2 Joules. The instantaneous power due to the force at the end of the interval is 120.932 Watts, and at the end of the first half of the interval it is 60.466 Watts. To find the force, we can use the equation F=ma, where a is the acceleration calculated to be 2.979 m/s^2.
AnkhUNC
[SOLVED] Power Problem

## Homework Statement

An initially stationary 2.9 kg object accelerates horizontally and uniformly to a speed of 14 m/s in 4.7 s. (a) In that 4.7 s interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?

## The Attempt at a Solution

Not really sure what's involved in this question. I need to find Power which is F*velocity but I don't know the force. Force = m*a. What I am given is the mass and speed is the velocity right? So I need to solve for a. Is this correct?

AnkhUNC said:
Force = m*a. What I am given is the mass and speed is the velocity right? So I need to solve for a. Is this correct?

Sounds good to me.

So finding a via v=v0+at I get a = 2.978723404. So for a it should be m*a = 8.638297872 but this is wrong. What did I do wrong :(?

AnkhUNC said:
So finding a via v=v0+at I get a = 2.978723404. So for a it should be m*a = 8.638297872 but this is wrong. What did I do wrong :(?

a Asks you for WORK. What is the definition of work done by a constant force?

Last edited:
Nono I meant a = acceleration. I get 8.638 N for the force.

I get the following:
a = 2.979; F = ma = 2.9*2.979 = 8.638 N

V1 = a*t/2 = 2.979*(4.7/2) = 7.000 m/s
V2 = 14.000 (given)

P1 = F*V1 = 8.638*7 = 60.466 W = (c)
P2 = 8.638*14 = 120.932 W = (b)

But it was (a) to be in Joules. However when I multiply my F in Newtons by meters (14*4.7) the answer isn't correct :( but (b) and (c) are.

Last edited:
W = KE = ½mV² = ½*2.9*14² = 284.2 J

## 1. What is the difference between work and instantaneous power?

The main difference between work and instantaneous power is the concept of time. Work refers to the amount of energy transferred when a force is applied to an object over a certain distance. It is calculated by multiplying the force by the distance traveled. Instantaneous power, on the other hand, is the rate at which work is being done at a specific moment in time. It is calculated by dividing the work done by the time it took to do it.

## 2. How do you solve a power problem using the work-energy theorem?

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. To solve a power problem using this theorem, you would first calculate the work done on the object by multiplying the force applied by the distance traveled. Then, you would calculate the change in the object's kinetic energy by subtracting its initial kinetic energy from its final kinetic energy. Finally, you can use the equation P = W/t to find the power, where P is power, W is work, and t is time.

## 3. Can you give an example of a power problem involving work and instantaneous power?

Yes, an example of a power problem involving work and instantaneous power could be a car accelerating from 0 to 60 miles per hour in 10 seconds. In this scenario, the work done on the car would be the force of the engine pushing the car forward multiplied by the distance it traveled. The instantaneous power would be the rate at which this work is being done, or the work divided by the time it took for the car to reach 60 miles per hour.

## 4. How can you calculate power if you know the force and velocity of an object?

If you know the force and velocity of an object, you can calculate the power using the formula P = Fv, where P is power, F is force, and v is velocity. This formula is commonly used in physics when dealing with objects in motion.

## 5. What are some real-world applications of solving power problems?

Solving power problems has many real-world applications, such as calculating the power output of a car or other vehicles, determining the power consumption of household appliances, and understanding the energy efficiency of different machines. This knowledge is also important in industries such as engineering and construction, where power calculations are necessary for designing and building structures and machinery.

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