A cart of mass 1.2 kg coasts....?

Question

A cart of mass 1.2 kg coasts 75 cm down an incline at 8 degrees with horizontal. Assume that frictional and other nongravitational forces parallel to the incline are negligible.
•What is the component of the cart's weight parallel to the incline?
•How much work does this force do as the cart rolls down the incline?
•Using the definition of kinetic energy determine the velocity of the cart after coasting the 75 cm, assuming its initial velocity to be zero.
•Using the definition of kinetic energy determine the velocity of the cart after coasting the 75 cm, assuming its initial velocity to be .45 m/s.

Answer 1

(a)wi=wsin(8)=mgsin(8)
w - weight
wi – component of weight along the plane
g – gravitational acceleration

wi=1.2 x 9.81 x sin(8)= 1.64 N

(b) W=Fs=wi s
W- work
F – force
s- displacement along which force F was applied.

W= mg sin(8) 0.75=
Or W=P=wh=mg sin(8) (same as W)
P - potential energy

W= 1.2 x 9.81 x sin(8) x 0.75=
W= 1.23 Joules

(c) P(top)=Ke(bottom)
P –potential energy
Ke – kinetic energy
Ke=0.5mv^2

v= sqrt(2Ke/m)
v= sqrt(2mg sin(8)/m)
v=sqrt(2gsin(8)
v= 1.65m/s

(d) V=v1+v2
V=1.65 + 0.45= 2.10m/s

I hope it helps