How to calculate work and power in a simple integral word problem?

[Jimbo57]

Homework Statement

A 100 lb bag of sand is lifted for 2 seconds at the rate of 4 feet per second . find the work done in lifting the bag if the sand leaks out at the rate of 1.5 lbs per second.

Homework Equations

The Attempt at a Solution

Here's my attempt at it:
work = force x distance
force = 100-1.5t
distance = 4t
work = 400t-6t²
Now I integrate for t=2 and t=0
∫(400t-6t²)dt
=200t²-2t³ for t=2 and t=0
=800-16=784ft-lb

Any mistakes?

 

[ehild]

The Attempt at a Solution

Here's my attempt at it:
work = force x distance
force = 100-1.5t
distance = 4t
work = 400t-6t²

This is not correct. Work =force times distance is valid only when the force is constant during the displacement.

Now I integrate for t=2 and t=0
∫(400t-6t²)dt
=200t²-2t³ for t=2 and t=0
=800-16=784ft-lb

Any mistakes?

If you integrate work with respect to time, you do not get work.

Use that power = force times velocity, and integrate the power over time.

ehild

 

[Jimbo57]

Ehild, thanks for pointing this out. I had a hunch that integrating work does not give you work, that should be common sense by now for me.

Correction:
power = force x velocity
force = 100-1.5t
velocity = 4
power = 400-6t
Now I integrate for t=2 and t=0
work = ∫(400-6t)dt
=400t-3t² for t=2 and t=0
=800-12=788ft-lb

 

[Dick]

Ehild, thanks for pointing this out. I had a hunch that integrating work does not give you work, that should be common sense by now for me.

Correction:
power = force x velocity
force = 100-1.5t
velocity = 4
power = 400-6t
Now I integrate for t=2 and t=0
work = ∫(400-6t)dt
=400t-3t² for t=2 and t=0
=800-12=788ft-lb

That looks much better.