Work and power from a 7.2 N force over three seconds

[danest]

Homework Statement

A force of 7.2 N acts on a 24 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.

Homework Equations

F = ma
P = dW/dT (derivative)
W= Fd(distance)

The Attempt at a Solution

For the solution I think that you would have to use F= ma to get the acceleration then use the constant acceleration formula to get the distance the body traveled. After that you get the work and divide it by the times?

 

[rl.bhat]

Hi danest, welcome to PF.
Your approach is correct. Try it.

 

[danest]

Hey thanks for the welcome :)

Okay I tried this out and I could only get the answer for the first part (the one second)

These are the things I got:
a = .3
distance = (1/2)(a)(t^2)

then i just inputed the a and for the t I put the 1 second and that allowed me to get the 1 second answer.
answer = 1.08

I then tried to same for parts two and three and then I divided the result by that amount of seconds and It was wrong.

For part d I am not sure on what to do. Would it be the same as the answer to the three second?

 

[rl.bhat]

Show the calculations for second and third second.
For d, find the velocity at the end of the third second. Then the instantaneous power P = F*v

 

[danest]

for the second and third I did this:
Second:
distance = (1/2)(.3)(2^2)
distance = .6
then I did F * d = 7.2 * .6 = 4.32 (I thought this would be the work done )

Third:
distance = (1/2)(.3)(3^2)
distance = 1.35
then I did F * d = 7.2 * 1.35 = 9.72

 

[rl.bhat]

The distance traveled in 2nd second is d2 - d1.
The distance traveled in 3rd second is d3 - d2.
You have already found d1, d2 and d3.
Now find the required distances and the work done.

 

[danest]

Thank you! I got the problem now.

Just in case anyone is wondering you could have also done this:

Integral from t2 to t1 of (F^2/m)t dt