Simple Integral Word Problem

  • Thread starter Jimbo57
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  • #1
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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?
 

Answers and Replies

  • #2
ehild
Homework Helper
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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
 
  • #3
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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
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
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.
 

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