How Much Work Does a Machine Perform Between 9s and 18s?

[pph1011]

Homework Statement

A machine does work at a rate given by , P=ct^2 where c= 18 W/s^2 and t is time.
Find the work done by the machine between 9.0s and 18s .

Homework Equations

P=w/t

The Attempt at a Solution

no ideas

 

[asleight]

Homework Statement

A machine does work at a rate given by , P=ct^2 where c= 18 W/s^2 and t is time.
Find the work done by the machine between 9.0s and 18s .

Homework Equations

P=w/t

The Attempt at a Solution

no ideas

Integrate with respect to time because:

$$P=\frac{dW}{dt}$$.

 

[thomate1]

I would approach this problem by first acknowledging that the given equation, P=ct^2, represents the power of the machine, not the work done. Power is the rate at which work is done, so in order to find the work done by the machine, we must use the formula P=w/t, where P is power, w is work, and t is time.

Next, I would substitute the given values into the formula. Since we are looking for the work done between 9.0s and 18s, we can set t=9.0s and t=18s respectively. This gives us two equations:

P1 = w/9.0s
P2 = w/18s

Since we know that P=ct^2, we can substitute that into the equations:

P1 = c(9.0s)^2
P2 = c(18s)^2

Finally, we can substitute the given value for c (18 W/s^2) into the equations and solve for w:

P1 = (18 W/s^2)(9.0s)^2 = 1458 J
P2 = (18 W/s^2)(18s)^2 = 5832 J

Therefore, the work done by the machine between 9.0s and 18s is 5832 J - 1458 J = 4374 J.

In conclusion, by using the correct formula and substituting the given values, we can find the work done by the machine between 9.0s and 18s. This approach demonstrates the importance of understanding the difference between power and work and using the appropriate equations in problem solving.