Can you help with this problem?

Question

A crate of mass 11.5 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100N parallel to the incline, which makes an angle of 21.0 degrees with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.05 m.

There are five questions to answer, but I need the final velocity in order to figure them out. I'm stuck. Can you help?

Ramin, Your answer is actually incorrect (*3.82 m/s*), but you gave me a jump start on how to figure it out. Thanks.

Answer 1

Second Newton's Law:
F-mg*Sin(b)-mg*Cos(b)*k=m*A
b=21 degrees
F=100 N, Sin(b)=0.35, Cos(b)=0.93, m=11.5 Kg, k=0.400
Therefore:
100-11.5*10*0.35-11.5*10*0.93*0.400=11.5*A
A=0.12 m/s^2
v(t)^2-v(t=0)^2=2*A*(x(t)-x(t=0))
v(t=0)=1.5 m/s, x(t)-x(t=0)=4.05 m
Therefore:
v(t)=4.79 m/s
End. Thank U... Ramin

Answer 2

Given: m=11.5kg
Therefore,w=mg=11.5g
Initial velocity, =1.50m/s
Coefficient of friction, u=0.400
Displacement of crate,s=4.05m
Applied force upward and parallel to incline=100N

force of friction, f=coeff*R where R is the force normal to the plane. R=11.5*9.8=112.7N.

f=.4*112.7
=45.1N

Net force, F=100-45.1
=54.9N

F=ma
54.9=11.5a
a=54.9/11.5
=4.77m/s^2

v^2-u^2=2as
v^2-1.5^2=2*4.77*4.05
v^2=38.6+2.25
=40.9
v=6.39m/s

Answer 3

In order to find out the final velocity, first of all you need to know either the acceleration or the time. Since time is difficult to work out, we can find the acceleration. I do not know How to write the free body diagram in this coloumn, but if you write one, then by newton's second law, we can write the equation
F-mgsina-f=ma -------------------------(1)
where, F=100N, m=mass of the crate,g=acceleration due to gravity, a=angle of the incline and f=frictional force.But we know that f=nmgcosa, where n=coefficient of kinetic friction.substituting we get,
F-mgsina-nmgcosa=ma=>
a=F-mgsina-nmgcosa/m.-----------------------(2)
Once we get the value of a , substitute in the equation of motion,
V^2=U^2+2as,-----------------------------------(3)
Where V=final velocity, that is to be found
U=Initial velocity, a=acceleration just found, S=the distance on the incline travelled, in this case 4.05m. (The free body diagram will clearly explain how to get the first equation.)

If you solve the equation (2), you get a=1.52m/s^2
if you sole, equation (3), you get,V=3.17m/s^2,with a negative sign.
The negative sign indicates that the velocity is slowing down.