Another one for ya.....?

Question

A 0.5-kg air-hockey puck is initially at rest. What will its kinetic energy be after a net force of 0.6 N acts on it for a distance of 2 m?
(a) 0.3 J
(b) 0.6 J
(c) 1.2 J
(d) 2.4 J

Answer 1

1.2j

Answer 2

I hope we can agree that Kinetic Energy, KE = 1/2 m*v*v. We have the mass but we do not have the velocity after it's been accelerated. So the question is: what is acceleration that occured when the puck was acted on by a force And what final velocity did that acceleration give to our puck.

First: Newton's second law. F = m*a.

Treat the left hand and right hand sides as two seperate equations. Remember you will have as many F = ma equations as you have axes in the coordinate system you choose - not important for this problem since you only have motion along one direction and are only interested in one direction.

F, F is the net force. Normally you would aim to find all the forces acting in the system along the direction you're interested but it's given to you here. F = .6N.

m*a, ask yourself is there an acceleration on the system, if there is, 'a' is non-zero. Then ask yourself if your system has a mass (a silly question, but a nice step to do until you understand what's going on better.) In our case, yes there is a mass and there is an acceleration that occurs on our puck. The mass of our system is .5 kg and the acceleration is unknown, but this law allows us to find the acceleration.

F = ma => .6N = .5 kg * a => 1.2 m/s/s = a

Once we know the acceleration that occured, we need a formula that will give us the final velocity of our puck. We know the initial velocity of the puck to be zero. The ideal equation then would probably have Initial velocity 'Vi', final velocity 'Vf' (what we want), acceleration 'a' and distance traveled 'x'. The equation we're looking for is:

Vf^2 = Vi^2 + 2*a*x
Vf^2 = 0 + 2*1.2m/s/s * 2m => Vf^2 = 4.8 (m/s)^2

Now that we know that Vf^2 is equal to 4.8 m/s we can find the final KE of our puck.

KE = 1/2*m*V^2 = 1/2 * (.5 kg) * (4.8 m/s) = 1.2 Kg *(m/s)^2 = 1.2 Joules

Answer 3

Since the puck is initially at rest, the kinetic energy will be the same as the work done on the puck, which is given by force X distance. In this case a force of 0.6 N acts for a distance of 2 m and thus the work done = KE = 1.2 J

Answer 4

first find its acceleration:
F = MA
A = F/M = 0.6 N / 0.5 kg = 1.2 m/s^2

so its velocity will be:
V^2 = 2ax
V = (sqrt)(2ax) = (sqrt)(2 * 1.2 m/s^2 * 2 m)
V = (sqrt)(4.8 m^2/s^2) = 2.19 m/s

so kinetic energy is:
E = (mv^2)(1/2)
E = (0.5 kg * 4.8 m^2/s^2)(1/2) = 1.2 J