# physics

posted by on .

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

• physics - ,

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
v2
= 2*1.757*0.73
= 2.565
v=sqrt(2.565)
=1.60 m/s