Problem in calculating work done?

In summary, the conversation discusses solving a question involving work done and integration. The attempt at a solution involves substituting values for F and ds, but it is incorrect to substitute "t^2+2t" for ds because ds represents a distance and cannot be directly replaced with a variable. Instead, the correct substitution is dx = (2t+2)dt.
  • #1

navneet9431

Gold Member
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Homework Statement


IMG_20180817_085052.jpg

See question number 3

Homework Equations


Work Done="integral" F*ds

The Attempt at a Solution


I tried to solve this question using integration,
IMG_20180817_085548.jpg

I have replaced F with "1" and ds with "t^2+2t".
So I am stuck in at that step.
Please help me differentiate it further or solve it further!
I will be thankful for any help!
 

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  • #2
In the integral it’s ##F*ds## which becomes ##F*dx## so you can’t just sub in the ##t^2+2t## term.

##dx/dt = 2t+2## and so ##dx = (2t+2)dt##
 
  • #3
jedishrfu said:
In the integral it’s ##F*ds## which becomes ##F*dx## so you can’t just sub in the ##t^2+2t## term.

##dx/dt = 2t+2## and so ##dx = (2t+2)dt##
Thanks for the reply!
But can you please explain why is it wrong to substitute "t^2+2t" in place of ds?
 
  • #4
navneet9431 said:
Thanks for the reply!
But can you please explain why is it wrong to substitute "t^2+2t" in place of ds?
You are not given ds (or dx)= t2+2t; you are given x=t2+2t. So what does dx equal?
 
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Likes navneet9431 and jedishrfu

1. What is work done in physics?

Work done in physics is a measure of the energy transferred to an object by means of a force acting on the object. It is defined as the product of the magnitude of the force and the displacement of the object in the direction of the force.

2. What is the formula for calculating work done?

The formula for calculating work done is W = F * d * cos(theta), where W is the work done, F is the magnitude of the force, d is the displacement of the object, and theta is the angle between the force and displacement vectors.

3. How do I calculate work done when the force is not constant?

When the force is not constant, work done can be calculated by finding the area under the force-displacement curve using integration. This method takes into account the varying force and displacement values throughout the motion.

4. Can work done be negative?

Yes, work done can be negative. This occurs when the force and displacement vectors are in opposite directions, resulting in the work done being in the opposite direction of the displacement. Negative work done means that energy is being taken away from the object.

5. How is work done related to power?

Work done and power are related through the equation P = W/t, where P is power, W is work done, and t is time. This means that the higher the power, the more work can be done in a given amount of time.

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