The force applied on a moving object with a mass of #3 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 [0,2 ] #?

Answer 1

I found: #W=14/3J#

Having a variable force I would use the work in integral form as: #W=int_(x_1)^(x_2)F(x)dx# along #x#. We have: #W=int_0^2(x^2+1)dx=# Solve it as usual: #=int_0^2(x^2+1)dx=x^3/3+x=# Apply the extremes: #=(2^3/3+2)-(0^3/3+0)=8/3+3=(8+6)/3=14/3# so we get: #W=14/3J#

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Answer 2

Olivia Askew

To calculate the work done, integrate the force function over the given interval [0, 2] with respect to x:

∫[0 to 2] (x^2 + 1) dx = [(1/3)x^3 + x] from 0 to 2

Plug in the upper and lower limits:

[(1/3)(2)^3 + 2] - [(1/3)(0)^3 + 0] = (8/3 + 2) - (0 + 0) = 14/3 units of work.

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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.