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
See:
http://www.jiskha.com/display.cgi?id=1244428188
To find the speed of the book at the end of the displacement, you can use the work-energy principle.
First, you correctly calculated the work done on the book as 13.91 J using the equation W = Fdcos(theta).
Now, let's apply the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy:
W = ΔK
Since the book begins with zero kinetic energy, we can rewrite this as:
W = Kf - Ki
Where Kf is the final kinetic energy and Ki is the initial kinetic energy.
Since the book starts from rest (zero initial kinetic energy), we can simplify the equation to:
W = Kf
Now, substituting the work value we calculated, W = 13.91 J.
We can also express the final kinetic energy as:
Kf = (1/2)mv^2
Where m is the mass of the book and v is its final speed.
Substituting the values given in the problem, mass (m) = 2.86 kg, and work (W) = 13.91 J, we get:
13.91 J = (1/2)(2.86 kg)v^2
Solving for v, we have:
v^2 = (2 * 13.91 J) / (2.86 kg)
v^2 = 27.82 J / 2.86 kg
v^2 = 9.72 m^2/s^2
Taking the square root of both sides, we find:
v = √9.72 m^2/s^2
v ≈ 3.12 m/s
Therefore, the speed of the book at the end of the displacement is approximately 3.12 m/s.