Plz help me Physics people!!?

Question

A 60.0 kg crate is attached to a weight by a cord that passes over a frictionless pulley. (1) If the coefficient of friction is 0.500, what weight will keep the crate moving up a 40 degrees incline at a constant speed? (2) If the cord is cut when the crate is at rest at the top of the incline, how far would the crate have slid by the time its speed reached 7.50 m/s?

Where do I even start? Can someone please explain this to me. I would be very, very thankful. Thanks.

Answer 1

a) Force equation is
W2=W1sin(40) + uW1cos(40)
W2=W1(sin(40) + u cos(40) )
W2=60.0 x 9.81(sin(40) + 0.500 cos(40))=
W2=604 N

b) V=at
S=0.5at^2 where t= V/a
and
a=F/m= [W1sin(40) - uW1cos(40) ]/m
a=g[sin(40) - u cos(40)]

t= V/a = 7.5/(9.81(sin(40) - 0.500 cos(40))=
t=7.5/ (2.54)=2.94 sec
S=0.5at^2
S=0.5 (2.54) (2.94)^2=11m