Physics Question Help?

Question

Pete slides a crate up a ramp at an angle of
27.6 degrees by exerting a 239 N force parallel to the
ramp. The crate moves at constant speed.
The cofficient of friction is 0.112.
How much work has been done against
gravity when the crate is raised a vertical
distance of 1.52 m? Answer in units of J.

Answer 1

Work (against gravity )= mgh

ballance of forces
f=F
f=umgcos(27.6)
m= F/ugcos(27.6)

Work (against gravity )= [F/ugcos(27.6) ]g h
Work (against gravity )= Fh/ [u cos(27.6) ]

Work (against gravity)= 239 x 1.52 / [ 0.112 cos(27.6)]=
Work (against gravity)= 3660 J

Answer 2

♥ work done is W=W1+W2, where work against friction is
W1=kN*L, k=0.112, N=mg*cos(b) is normal component of weight of the crate, b=27.6°, L is distance up the ramp,
work against gravity is
W2=mgh, h=L*sin(b)=1.52m, while total work is
W=F*L, F=239 N;
♠ thus F*L = k* mg*cos(b)*L + mg* L*sin(b), hence weight of the crate is
mg = F/(k*cos(b) + sin(b));
♣ therefore W2=mg*h = F*h/(k*cos(b) + sin(b))=
= 239*1.52 /(0.112*cos(27.6) + sin(27.6)) =645.77 J;