Question

The force applied against an object moving horizontally on a linear path is described by `F(x)=x^2-3x + 3`. By how much does the object's kinetic energy change as the object moves from `x in [ 0 , 1 ]`?

Answer 1

Newton's second law of motion:

`F=m*a`

Definitions of acceleration and velocity:

`a=(du)/dt`

`u=(dx)/dt`

Kinetic energy:

`K=m*u^2/2`

Answer is:

`ΔK=11/6` `kg*m^2/s^2`

Explanation:

Newton's second law of motion:

`F=m*a`

`x^2-3x+3=m*a`

Substituting `a=(du)/dt` does not help with the equation, since `F` isn't given as a function of `t` but as a function of `x` However:

`a=(du)/dt=(du)/dt*(dx)/dx=(dx)/dt*(du)/dx`

But `(dx)/dt=u` so:

`a=(dx)/dt*(du)/dx=u*(du)/dx`

Substituting into the equation we have, we have a differential equation:

`x^2-3x+3=m*u(du)/dx`

`(x^2-3x+3)dx=m*udu`

`int_(x_1)^(x_2)(x^2-3x+3)dx=int_(u_1)^(u_2)m*udu`

The two speeds are unknown but the positions `x` are known. Also, mass is constant:

`int_(0)^(1)(x^2-3x+3)dx=m*int_(u_1)^(u_2)udu`

`[x^3/3-3x^2/2+3x]_0^1=m*[u^2/2]_(u_1)^(u_2)`

`(1^3/3-3*1^2/2+3*1)-(0^3/3-3*0^2/2+3*0)=m*(u_2^2/2-u_1^2/2)`

`11/6=m*u_2^2/2-m*u_2^2/2`

But `K=m*u^2/2`

`11/6=K_2-K_1`

`ΔK=11/6` `kg*m^2/s^2`

Note : the units are `kg*m^2/s^2` only if the distances given `(x in[0,1])` are in metres.