How Do You Calculate Final Speed from Work Done on an Object?

  • Thread starter Thread starter alexphysics
  • Start date Start date
  • Tags Tags
    Final Speed
alexphysics
Messages
3
Reaction score
0
Finding Final Speed Using Work

A person pushes with a horizonal 25N force over a distance of 1.2m on a 4.5kg crate, intially at rest on a frictionless surface. What is the final speed of the crate?

heres my data list..

F=25N
delta d=1.2m
m=4.5kg
v1=0
v2=?
theta=0 degrees

i used the W=fxdeltadxcos(theta) formula to find out the work done was 30J.. but once i have work i don't no what to find next to find the final speed..
 
Last edited:
Physics news on Phys.org
There are two ways of approach. One is using the constant acceleration equation

(v2)^2=(v1)^2+2as

where a is the acceleration of the crate and s is the displacement caused by the acceleration, i.e.

v1 = 0, a = F/m (Newton's Second Law) = 25/4.5 = 50/9, s = 1.2, so

v2 = sqrt (2.4*50/9) = sqrt (120/9 ) = sqrt (120)/3

The other is to find the work done by the force = F.d = 25*1.2 = 30J
and then say this is the additional kinetic energy given to the crate. Since the crate was at rest, we can say

0.5*m*(v2)^2 = 30
v2 = sqrt ( 60/4.5 ) = sqrt (120/9) = sqrt (120)/3
 
What is the work done by the force converted into? (If it is moving what kind of energy does it have?)
 
Energy by virtue of its motion is called kinetic energy, and kinetic energy is given by the formula K.E. = 0.5mv^2 where m is the mass of the body and v is the velocity.
 

Similar threads

  • · Replies 56 ·
2
Replies
56
Views
5K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
10
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 2 ·
Replies
2
Views
8K