A 1.75 kg puck is placed into a spring loaded launcher on a frictionless horizon

Question

A 1.75 kg puck is placed into a spring loaded launcher on a frictionless horizontal tabletop. The spring has a spring constant of 250 N/m and is compressed 0.40 m.


a) What is the initial energy?
b) What is the speed of the puck after it leaves the launcher?
c) The puck then slides onto a 0.80 m long section of the table that is sticky and has a coefficient of kinetic friction of 0.45 . How fast is the puck going after it leaves the sticky part of the table?
d) Shortly after that, the puck slides off the end of the table. If the table is 0.98 m tall, how fast is it going when it hits the floor (do not use kinematics here).

Explanation / Answer

a. Hooke's law gives that:

Espring = .5 k (x)2 = .5 (250 N/m) (0.40 m)2 = 20 N m = 20 J

b. We know that energy is conserved in the problem, so

Ekinetic = Espring

.5 M v2 = 20 J

M v2 = 40 J

(1.75 kg) v2 = 40 J

v = sqrt(40/1.75) = 4.78 m/s

c. While the puck slides along the sticky portion, there is a force:

Ffriction = Fnormal = 0.45 * M * 9.8 m/s2 = M * 4.41 m/s2

a = Ffriction / M = 4.41 m/s2

We can now plug this into one of our equations of motion, keeping in mind its a negative acceleration:

v12 = v02 + 2a(x)

v12 = (4.78 m/s)2 - 2 (4.41 m/s2) (0.80 m) = 15.8 m2/s2

v1 = 3.97 m/s

d. We know the puck's x-velocity should be the same when it hits the ground (since no forcers in the x-direction). We use conservation of energy to find the y-component of the velocity:

Epotential = Ekinetic

m g y = .5 m vy2

(1.75 kg) (9.8 m/s2) (0.98 m) = .5 (1.75 kg) vy2

vy = sqrt(((1.75 kg) (9.8 m/s2) (0.98 m)) / (.5 (1.75 kg))) = 4.38 m/s

v = sqrt(vx2 + vy2) = sqrt((3.97 m/s)2 + (4.38 m/s)2) = 5.91 m/s

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