• Support PF! Buy your school textbooks, materials and every day products Here!

Find work done given time, acceleration, and mass

  • Thread starter daltomagne
  • Start date
  • #1
a box with a mass of 6.0 kg is accelerated from rest by a force across a floor at a rate of
2.2 m/s^2 for 5.9 s. Find the net work done on the box

m=6.0kg
a=2.2m/s^2
t=5.9 s
a=0 degrees

i know W=Fdcos(a)
and F=ma


so,
F=6.0kg*2.2m/s^2=13.2N
W=13.2N*d*cos0=13.2N*d

I am a little unsure as how to find d though. would it be one of the kinematic equations?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi daltomagne! :smile:

(try using the X2 icon just above the Reply box :wink:)
a box with a mass of 6.0 kg is accelerated from rest by a force across a floor at a rate of
2.2 m/s^2 for 5.9 s. Find the net work done on the box

I am a little unsure as how to find d though. would it be one of the kinematic equations?
That's right. :smile:

You can either use one of the standard constant acceleration equations to find d, or you can use another of them to find vf, and then apply the work-energy theorem ( work done = change in mechanical energy).
 
  • #3
so i'm thinking v=vi+at and that gives me vi=-12.98m/s
and
xf=xi+vit+1/2at2?
but that gives me a value for x=-38.3 m?

so something isn't adding up
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
251
where's the contradiction? :confused:

one figure is speed, the other is distance, they should both give you the same work. :wink:
 

Related Threads on Find work done given time, acceleration, and mass

Replies
6
Views
1K
Replies
9
Views
1K
Replies
8
Views
960
Replies
1
Views
1K
Replies
14
Views
618
Replies
10
Views
889
Replies
4
Views
1K
Replies
1
Views
670
Replies
18
Views
3K
Replies
2
Views
2K
Top