do i use Fd=1/2mv(squared)+mgh for this problem?

Question

You are pulling your sister on a sled to the top of a 15.0 m high, frictionless hill with a 10.0° incline. Your sister and the sled have a total mass of 50.0 kg. You pull the sled, starting from rest, with a constant force of 127 N at an angle of 45.0° to the hill. If you pull from the bottom to the top, what will the speed of the sled be when you reach the top?

Answer 1

No, not required.
The component of the force along the hill = 127 cos 45

Net force in that direction = 127 cos 45 - 50 sin 10

127 cos 45 - 50 sin 10 = 50 X a

a = (127 cos 45 - 50 sin 10) / 50

This will give you "a"

Now, initial velocity is 0
you can calculate the distance to travelled along the hill,
s = 15 / sin 10

v^2 = 2as

Solve for v

Answer 2

No
F=MA
A=F/M
The hill as described is a straight line rather than a more realistic shape. Level on the top and curved with different slopes at different heights.
This simplified problem allows you to determine the component of gravity that is acting along the surface of the hill
F= sin(10 degrees)*g*mass=.174*9.8*50=85.1 newtons
The component of the force being applied by my pulling on the sled which is acting along the surface is
F2=sin(45 degrees)*127=.707*127=89.8 newtons

these two forces are in opposite directions so the net force is the difference between them.
Fnet=F2-F=89.8-85.1=4.7 newtons

a 10 degree slope with a 15 meter difference in altitude is
length=height/sin(10 degrees)=15/.174=86.4 meters

From D=1/2AT^2
and V=AT
and F=MA
A=Fnet/M
D=1/2*V^2/A
V^2=2*D*A=2*D*Fnet/M
V=Sqrt(2*D*F/M)=Sqrt(2*86.4*4.7/50)

V=4 meters/second