Question

The force applied against a moving object travelling on a linear path is given by `F(x)= sinx + 1`. How much work would it take to move the object over `x in [ 0,pi/3]`?

Answer 1

3.08 units

Explanation:

Work done W= `F.ds` = `m*(dv)/dt*ds`
= `m*dv(ds)/dt` = `m (dv)^2`
Given equation,
F = `sin x + 1`
Or, `mv (dv)/dx = sin x + 1`
Or, `mv dv = sin x dx + dx`
Integrating we get,
`m((v1)^2 - (v2)^2)= (-cos x + x)/2`
Putting the interval we get,
`m dv ^2 = (-(1/2)+1+(pi/3))*2` i.e 3.08 units

Answer 2

The work is `=1.55J`

Explanation:

The equation is

`"Work"="Force"xx"distance"`

Here,

The force is `F(x)=sinx+1`

Therefore,

`DeltaW=FxxDeltax`

`dW=(sinx+1)dx`

Integrating both sides

`W=int_0^(pi/3)(sinx+1)dx`

`=[-cosx+x]_0^(pi/3)`

`=(-cos(pi/3)+pi/3)-(-cos(0)+0)`

`=-1/2+pi/3+1`

`=1/2+pi/3`

`=1.55J`