physics
posted by mary .
The question is: a horizontal force of magnitude 22.0 N is applied to a 2.86 kg book as the book slides a distance d = 0.73 m up a frictionless ramp at angle θ = 30°. The book begins with zero kinetic energy. What is its speed at the end of the displacement?
I did W=fdcos(theta) and found the amount of work to be 13.91J.
I then thought maybe you use the equation K=(1/2)mv^2
and rewrite as W=(1/2)mv^2
in which case v=3.12m/s
but this was wrong.
any guidance is appreciated! THANKS

m=2.86 kg
g=9.81 m/s/s
θ=30 degrees
F=22 N horizontal
d=0.73 m (displacement)
μ=0 (coefficient of friction)
The weight of the mass can be resolved into two components,
Normal to incline = mg cos(θ)
along incline (downwards) = mg sin(θ)
The horizontal force can be resolved into two components
Normal to incline = F sin(θ)
along incline (upwards) = F cos(θ)
Net force pushing mass up incline
= F cos(θ)  mg cos(θ)
= 22 cos(30)  2.86*9.81 sin(30)
= 19.053  14.028
= 5.024 N up incline
Net acceleration up incline
= 5.024 / 2.86
= 1.757 m/s/s
initial velocity = 0
final velocity = v m/s
distance = d = 0.73 m
v^{2}  0^{2} = 2ad
v^{2}
= 2*1.757*0.73
= 2.565
v=sqrt(2.565)
=1.60 m/s