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The force applied on a moving object with a mass of #5 kg # on a linear path is given by #F(x)=x^2+1 #. How much work would it take to move the object over #x in [1,4 ] #?

Answer 1

Olivia Askew

I got: #24J#

Since the force is variable, I would employ the integral form of work as follows:

#W=int_(x_1)^(x_2)F(x)dx=#

In your situation:

#W=int_1^4(x^2+1)dx=x^3/3+x|_1^4=# #=(4^3/3+4)-(1^3/3+1)=(64+12-1-3)/3=24J#

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Answer 2

Carson Alexander

To calculate the work done, integrate the force function F(x) over the given interval [1,4]:

[ \text{Work} = \int_{1}^{4} (x^2 + 1) ,dx ]

[ \text{Work} = \left[\frac{1}{3}x^3 + x\right]_{1}^{4} ]

[ \text{Work} = \left(\frac{1}{3}(4)^3 + 4\right) - \left(\frac{1}{3}(1)^3 + 1\right) ]

[ \text{Work} = \frac{64}{3} + 4 - \frac{1}{3} - 1 ]

[ \text{Work} = \frac{61}{3} \text{ Joules} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.