Question

The force applied against a moving object travelling on a linear path is given by `F(x)=x+2e^x`. How much work would it take to move the object over `x in [0,2 ]`?

Answer 1

Work Done `=2e^2or approx 14.78J`

Explanation:

Work done by a Force `vecF` when it moves the object through a distance `vecdS` is

`dW=vecF.vecdS=|vecF|.|vecdS|.cos theta`

Where `theta` is the angle between the two vectors

As the Force is being applied on the object moving in a linear path the angle between the two is `0`, an therefore `costheta=1`

`dW=|vecF|.|vecdS|.`

When the force `F(x)` moves through a small distance `dx`
`dW=F(x)dx`
Total work done to move the object from `x in [0,2]` is integral of `dW`over limits `0` to `2`.

`W=int_0^2(x+2e^x)dx`
`=x^2//2+2e^x+C|_0^2`, where C is constant of integration.
`=(2^2//2+2e^2+C)-(0^2//2+2e^0+C)`
`=(cancel (color (red)2)+2e^2+cancel C)-((cancel (color (red)2)+cancelC)`
`=2e^2`