How Fast Does a 2100kg Block Travel After 3900J of Work?

[xregina12]

A machine pushes a 2100kg block from rest to a speed of v, doing 3900 J of work. During this time the car moves 10 meters. Neglecting friction between the floor and the block, what is v?
Work=3900J
d=10 meters
m=2100kg

KE=1/2 x m x v^2
3900=1/2 x 2100 x v ^2
v^2= 3900 x 2 / 2100
v = 1.93.

This is how I approached the problem but I'm not sure if it's right. I'm not sure if 3900 J's given equal the kinetic energy even though I set it equal to the KE. Can someone let me know if this is right? Thanks.

 

[Rake-MC]

I'm afraid that is incorrect. There is a total of 3900 J 'expended' in this system in the given time. However there is kinetic energy and work done.

K_e = \frac{1}{2}mv^2
W = F\Delta d

Here is a starter on how to do it, think of velocity as distance divided by time.
You know that acceleration is distance/time/time.
You know Force = acceleration * mass.
You can get the equation solely in terms of mass, distance and time. Mass and distance you are given. Solve for time, solve equation.

 

[PhanthomJay]

A machine pushes a 2100kg block from rest to a speed of v, doing 3900 J of work. During this time the car moves 10 meters. Neglecting friction between the floor and the block, what is v?
Work=3900J
d=10 meters
m=2100kg

KE=1/2 x m x v^2
3900=1/2 x 2100 x v ^2
v^2= 3900 x 2 / 2100
v = 1.93.

This is how I approached the problem but I'm not sure if it's right. I'm not sure if 3900 J's given equal the kinetic energy even though I set it equal to the KE. Can someone let me know if this is right? Thanks.

Yes, you are correct. Per the work-energy theorem, the total work done on an object is equal to its change in KE. Since the only work being done is by the truck, the 3900J is the total work. Just be mindful when and when not to use this equation.

 

[xregina12]

thanks!

 

[Rake-MC]

So the 3900J done IS the F\Delta d ?
Well then what I said before was wrong.

 

[PhanthomJay]

So the 3900J done IS the F\Delta d ?
Well then what I said before was wrong.

Not really, it leads to the same result. The work done is 3900J=Fd, the force applied is thus 390N; the acceleration is therefore 390/2100 = .186m/sec^2, and the time taken to travel the 10m per s= 1/2at^2 is = 10.38 sec, and v=at = 1.93m/s (whew!)