Question

Question #7a2db

Answer 1

`v_x = 7.60` `"m/s"`

Explanation:

We're asked to find the speed of the wagon after it travels a distance of `5.0` `"m"`.

To to this, we need to find the net horizontal force `sumF_x` acting on the wagon.

We're given the coefficient of kinetic friction `mu_k` is `0.6`, and we have the equation

`f_k = mu_kn`

where

• `f_k` is the magnitude of the kinetic friction force (opposes motion)

• `n` is the magnitude of the normal force exerted by the surface on the wagon

The child is pulling the wagon at an angle `30^"o"` from the horizontal, so the magnitude of the normal force is

`n = overbrace(mg)^"weight" - overbrace(Fsin30^"o")^"vertical component of applied force" = sumF_y`

The magnitude of its weight `mg` is

`(15color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 147` `"N"`

So we have

`n = 147color(white)(l)"N" - (150color(white)(l)"N")sin30^"o" = color(red)(72.2` `color(red)("N"`

So the magnitude of the friction force is

`f_k = (0.6)(color(red)(72.2color(white)(l)"N")) = color(green)(43.3` `color(green)("N"`

The net horizontal force `sumF_x` is

`sumF_x = overbrace(Fcos30^"o")^"horizontal component of applied force" - overbrace(f_k)^"opposing friction force"`

`= (150color(white)(l)"N")cos30^"o" - color(green)(43.3color(white)(l)"N") = color(purple)(86.6` `color(purple)("N"`

Now that we have the net horizontal force acting on the wagon, we can use Newton's second law to find its acceleration:

`sumF_x = ma_x`

`a_x = (sumF_x)/m = color(purple)(86.6color(white)(l)"N")/(15color(white)(l)"kg") = color(orange)(5.77` `color(orange)("m/s"^2`

We finally have the object's horizontal acceleration, so we can now use kinematics equations to find the speed after it covers `5.0` `"m"`:

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

In this case,

• `v_x` is the speed (we're trying to find this)

• `v_(0x)` is the initial speed, which is `0` since it started from rest

• `a_x` is `color(orange)(5.77` `color(orange)("m/s"^2`

• `Deltax` is the distance covered, which will be `5.0` `"m"`

Plugging in known values, we have

`(v_x)^2 = 0 + 2(color(orange)(5.77color(white)(l)"m/s"^2))(5.0color(white)(l)"m")`

`v_x = sqrt(2(color(orange)(5.77color(white)(l)"m/s"^2))(5.0color(white)(l)"m")) = color(blue)(7.60` `color(blue)("m/s"`