Question

An object with a mass of `1 kg` is pushed along a linear path with a kinetic friction coefficient of `u_k(x)= 3x^2+12`. How much work would it take to move the object over #x in [1, 3], where x is in meters?

Answer 1

Work `W=490" "`Joules

Explanation:

Given:
mass `m=1 " ""kg"`
Variable kinetic friction coefficient `mu_k(x)=3x^2+12`

Let `x` the distance moved by the object

Let `F_k` the force required to move the object
Let `F_n` the normal force exerted by the surface on the object
`F_n=m*g`

`F_k=mu_k*F_n`
`F_k=(3x^2+12)*m*g`

The amount of work

`dW=F_k*dx`

`W=int_1^3 F_k*dx`

`W=int_1^3 (3x^2+12)*m*g*dx`

`W=mg*int_1^3 (3x^2+12)*dx`

`W=mg[(x^3+12x)]_1^3`

`W=50*mg`

`W=50*(1)*(9.8)`

`W=490" "`Joules

God bless...I hope the explanation is useful