Calculate Final Kinetic Energy of 2.0 kg Object | 30N & 15N

[wikidrox]

I need some assistance in answering this:

A force of 30N accelerates a 2.0 kg object from rest for a distance of 3.0 m along a frictionless horizontal surface. The force then changes to 15N and acts for an additional 2.0 m. What is the final kinetic energy of the object?

Would the final Kinetic Energy just be the kinetic energy of both situations added together?

 

[Doc Al]

work-energy theorem

Find the work done by each force. The total work done will equal the change in KE.

 

[BobG]

There's also a long way to do this without using integrals - find out how fast you're going after 5 meters and plug into your kinetic energy formula (1/2 mv^2).

The long way, you have 2 problems:

Problem 1:

From your first force, you can find acceleration resulting from that force.

Starting from your initial position (zero) and from your initial velocity (zero) and adding in your acceleration, you can find your final position. In this problem, you have your final position (3 meters) and need to solve for time.

Starting from your initial velocity (zero) and adding in acceleration over a given amount of time (which is why you needed to do the previous problem), you can find your final velocity.

If you wanted, you could determine how much kinetic energy the object has after 3 meters (but, the problem doesn't ask for this).

Problem 2:

From your second force, you can find acceleration resulting from that force.

Starting from your initial position (final position from problem #1 or 3 meters), and from your initial velocity (final velocity from problem #1), and adding in your acceleration, you can find your final position. In this case, your final position is (3 meters + 2 meters) and you need to solve for time.

Starting from your initial velocity (final velocity from problem #1) and adding in your acceleration over a given amount of time (acc and time from prob #2) you can get your final velocity.

Now, using your final velocity, you can find your kinetic energy.

 

[Doc Al]

There's also a long way to do this without using integrals - find out how fast you're going after 5 meters and plug into your kinetic energy formula (1/2 mv^2).

...

Yikes, that's the long way alright.

Finding the work done by each force by direct calculation is trivial--No need for any integrals.

But, the more ways you can do a problem, the better your understanding.

 

[BobG]

Yeah, you're right.

But, then, I'll go to any length to avoid work

 

[Doc Al]

Good one!