Question

How much work would it take to horizontally accelerate an object with a mass of `4 kg` to `2 m/s` on a surface with a kinetic friction coefficient of `3`?

Answer 1

`W = 8` `"J"`

Explanation:

We're asked to find the work necessary to accelerate a `4`-`"kg"` object horizontally from rest to `2` `"m/s"` on a surface where the coefficient of kinetic friction is `3`.

To do this, we'll use the equation

`f_k = mu_kn`

where

• `f_k` is the magnitude of the retarding friction force

• `mu_k` is the coefficient of kinetic friction (`3`)

• `n` is the magnitude of the normal force, which since the plane is horizontal is equal to

`n = mg = (4color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 39.24` `"N"`

The kinetic friction force is thus

`f_k = (3)(39.24color(white)(l)"N") = 118` `"N"`

This means a force of at least `118` `"N"` is needed to keep the object accelerating. The force doesn't matter, because the larger the force, the shorter the displacement once the object reaches `2` `"m/s"`.

The work will be the same regardless of the force here; let's say the applied force is `119` `"N"`, so the net force is

`119` `"N"` `-118` `"N"` `= 1` `"N"`

The magnitude of the acceleration is

`a_x = (sumF_x)/m = (1color(white)(l)"N")/(4color(white)(l)"kg") = 0.25` `"m/s"^2`

Using kinematics, we can find the displacement `Deltax` after it reaches `2` `"m/s"`, using the equation

`(v_x)^2 = (v_(0x))^2 + 2a_x(Deltax)`

It starts from rest, so we have

`(2color(white)(l)"m/s")^2 = 0 + 2(0.25color(white)(l)"m/s"^2)(Deltax)`

`Deltax = 8` `"m"`

Using

`W = Fs`

we have

`W = (1color(white)(l)"N")(8color(white)(l)"m") = color(red)(8` `color(red)("N"·"m"` `= color(red)(8` `color(red)("J"`