A 96 kg crate starting from rest is pulled across a floor awith a constant horizontal force of 350N. For the first 15m the floor is frictionless and for the next 15m the coefficient of friction is .25. What is the final speed of the crate?

Question

anonymous · Answer

This uses the law of conservation of energy.

anonymous · Answer

I know that the conservative force is at point 1, by the horizontal force of 350N, I also know that the nonconservative force is at the second point, that being the coefficient of kinetic friction (I'm assuming since the crate is moving at this point) .25

anonymous · Answer

Since friction is moving in the opposite direction, it would be -.

anonymous · Answer

My first attempt at solvingthis problem is to find the total net work, that is, the amount of work done by the crate at point 1 and the work done by the crate with friction at point two.

anonymous · Answer

$$mu _{k}F _{N}= 96kg*9.80m/s^2*.25=-235N$$
NC means nonconservative work 
$$W _{NC}=235*15m=-3525J$$
I'm not sure if I've done this part correct for the work done at point one:

$$W _{C}= 320N*15m=5250J$$

anonymous · Answer

$$W _{net}=5250J-3525J=1725J$$
I think this is right so far

anonymous · Answer

Then from here I need a formula to find the velocity of the speed, I'm stuck at this point...

anonymous · Answer

|dw:1335234854664:dw|
F=350N
fk=$$mu$$k*m*g=0.25*96*9.8=235.2 N
S1=15m
S2=15 m

#) For the first 15 m:
$$\sum_{}^{}F=m \times a$$
350N= 96 kg x a
a= 350/96 m/s^2
$$Vt1^{2}=Vo ^{2}+2.a.S1$$
              =0+2*(350/96)*15
               = 875/8 m^2/s^2

#) For the second 15 m:
$$\sum_{}^{}F=m \times a$$
F-fk=m x a
350N-235.2=96kg x a
a= 287/240 m/s^2
$$Vt2^{2}=Vt1 ^{2}+2.a.S2$$
              = 875/8 + 2* (287/240)*15
              = 581/4 m^2/s^2
$$Vt2=\sqrt{581/4}=12,05  m/s ^{2}$$