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?

Question

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

Answer 1

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 - 0² = 2ad
v²
= 2*1.757*0.73
= 2.565
v=sqrt(2.565)
=1.60 m/s

Answer 2

To find the speed of the book at the end of the displacement, you'll need to apply the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

In this case, the work done on the book is given by the equation W = fdcos(theta), as you correctly calculated. Therefore, the work done is 13.91 J.

But the work done on the book is not equal to its kinetic energy. The work done is equal to the change in kinetic energy, which means the final kinetic energy minus the initial kinetic energy. Since the book begins with zero kinetic energy, the equation becomes:

13.91 J = (1/2)mv^2 - 0

Simplifying the equation, you have:

13.91 J = (1/2)(2.86 kg)v^2

Now you can solve for the velocity. Rearranging the equation gives:

v^2 = (2 * 13.91 J) / (2.86 kg)

v^2 = 27.82 J / 2.86 kg

v^2 = 9.715 m^2/s^2

Taking the square root of both sides, you find:

v ≈ 3.12 m/s

So the speed of the book at the end of the displacement is approximately 3.12 m/s.