Question

A student pushes a 25 kg crate which is initially at rest with a force of 160 N over a distance of 15 m. If there is 75 N of friction, what is the final speed of the crate?

`∆E_k = F_"net"∆d`

`v = sqrt((2F_"net"∆d)/m)`

(How do I find Fnet?)

Answer 1

`"10. m s"^(-1)`

Explanation:

Ok, so you know that the work done by a net force that acts on an object is equal to the change in that object's kinetic energy.

`color(blue)(ul(color(black)(W = DeltaK_E)))`

If we take `F_"net"` to be the net force and `d` the displacement of the object that results from applying the net force, we can say that

`W = F * d * cosalpha`

In your case, the force acts in the direction of the displacement, so

`alpha = 0^@ implies cos alpha = 1`

Similarly, if we take `m` to be the mass of the object, `v_i` the initial velocity of the object, and `v_f` the final velocity of the object, we can say that

`DeltaK_e = 1/2 m * v_f^2 - 1/2 * m * v_i^2`

In your case, `v_i = "0 m s"^(-1)`, so

`DeltaK_E = 1/2 * m * v_f^2`

This means that you have

`F_"net" * d = 1/2 * m * v_f^2`

which, as you know, gets you

`v_f = sqrt( (2 * F_"net" * d)/m)" " " "color(orange)("(*)")`

Now, the net force is simply the sum of all forces that act on the object, with the added mention that it is the vector sum, i.e. that it takes into account the magnitudes and the directions of the forces.

In your case, you know that the force `F = "160 N"` is acting to accelerate the object and that the force of friction is opposing `F`.

This means that the net force will be

`F_"net" = F - F_"friction"`

`F_"net" = "160 N" - "75 N" = "85 N"`

Keep in mind that the net force acts in the same direction as `F` (think vectors here).

This means that you have

`v_f = sqrt( (2 * 85 color(red)(cancel(color(black)("kg"))) "m s"^(-2) * "15 m")/(25color(red)(cancel(color(black)("kg"))))) = color(darkgreen)(ul(color(black)("10. m s"^(-1))))`

The answer is rounded to two sig figs.