Find the work done on the object?

Question

A single force acts on a 3.0 kg particle-like object whose position is given by x=3.0t-4.0t^2+1.0t^3, with x in meters and t in seconds. Find the work done on the object by the force from t=0 to t=4.0 s. 

Okay, so I did the integration. I got (t^4/4-4t^3/3+3t^2/2) from 0 to 4 and the calculator says -32/15. But the answer is 5.3x10^2 J. How do I get there?

? · Accepted Answer

Work done (in Joules) = (force on the object in Newton) * (displacement of the object)

Value of the force is given in the question in kg. Convert it Newton. 
Force in Newton = (force in kg)*9.81 
= 3*9.81 
=29.43 Newton

 Value of the displacement is not given but can be found by substituting value of t in the equation for the position. 

Equation of position is: x=3.0t-4.0t^2+1.0t^3
Position at t = 0 s is: x=3.0*0-4.0*0^2+1.0*0^3 = 0
Position at t = 4 s is: x=3.0*4-4.0*4^2+1.0*4^3 = 0
= 12-64+64 = 12 meters.
Displacement from t=0 to t=4 is 12-0 meters = 12 meters.

Now you have values of both the parameters for calculation of the work done: force (= 29.43 Newton) and displacement (=12 meter).
Work done (in Joules) is (force on the object in Newton) * (displacement of the object)
= 29.43*12 
= 353.16
= 3.5x10^2 Joules.

This is not matching with answer you are expecting. Either the answer is wrong or there is some typing error in the question. There is no mistake in the method and calculations in my answer.

? · Answer

I don't think you have to do integration.
Work is defined as force multiply by distance.
Work from t=0 to t=4 is W = W(4)-W(0)

Here, the force is function of time. F = m*a.
a can be obtained from second derivative of x.
we will get a= 6*t - 8

Then we can say F=3*(6t-8) = 18t -24

W=F*x
So, work in the function of time is:

W(t)= (18t -24)*(3.0t-4.0t^2+1.0t^3)

Work done from t=0 to t=4 is

W = W(4)-W(0)
W= 576 - 0
W= 576 Joule

I hope it helps

Anonymous · Answer

Acceleration is defined as the second derivative of space over a time to the square t.

So if you want the acceleration you need do d^2x/dt^2 = you then get acceleration and multiply by the mass to get the force. I suppose acceleration will be something like x' = 3 - 8t + 3t^2, sorry then you need to do the 2nd derivative so x'' = 6t - 8 there you can get the acceleration from time t0 and t4. But I don't remember about physics, it was long ago. I would then substitute the whole thing in the force formula, by multiplying it by the mass and then integrate it with x to get the Work.

A work is the force that acts on the particle of 3 kilos to make it move of an acceleration and a final space of x.

I don't know if this is correct, it's been a long time.

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Find the work done on the object?

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