# 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?)

(How do I find Fnet?)

##### 1 Answer

#### Answer:

#### 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 *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

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

In your case,

#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 **opposing**

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

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**.