A hockey puck of mass 0.228 kg si sliding across level ice. There is a force of friction of 0.057 N acting opposite the direction of motion of the puck while it is sliding.
a) what is the acceleration of the puck while it is sliding?
b)if it initially has a speed of 10.0 m/sec, how long will it take the friction to stop the puck, and how far will it slide in this time?
F = m A
so
A = -.057/.228
v = Vi + A t
0 = 10 - (.057/.228)t
t = 2.28/.057
d = 10 t -(.057/.456) t^2
or for d just use average speed during stop
5 * t
To find the acceleration of the puck while it is sliding, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
a) The force acting on the puck is the force of friction, which is given as 0.057 N. The mass of the puck is 0.228 kg.
Using Newton's second law:
Net force = mass * acceleration
Rearranging the equation:
Acceleration = Net force / mass
Substituting the given values:
Acceleration = 0.057 N / 0.228 kg
Calculating the acceleration:
Acceleration = 0.250 m/s²
Therefore, the acceleration of the puck while it is sliding is 0.250 m/s².
b) To find the time it takes for the friction to stop the puck and the distance it will slide in this time, we can use the equation of motion:
v = u + at
where:
v = final velocity (which is 0 m/s because the friction stops the puck)
u = initial velocity
a = acceleration of the puck (which is -0.250 m/s² because it acts opposite to the direction of motion)
t = time taken
We need to find the time taken, so we can rearrange the equation:
t = (v - u) / a
Substituting the given values:
u = 10.0 m/s
a = -0.250 m/s²
v = 0 m/s
Calculating the time taken:
t = (0 - 10.0) m/s / -0.250 m/s²
t = -40.0 s / -0.250 m/s²
t = 160 s
Therefore, it will take 160 seconds for the friction to stop the puck.
To find the distance it will slide in this time, we can use the equation of motion:
s = ut + (1/2)at²
where:
s = distance
u = initial velocity
t = time taken
a = acceleration
Substituting the given values:
u = 10.0 m/s
t = 160 s
a = -0.250 m/s²
Calculating the distance:
s = (10.0 m/s)(160 s) + (1/2)(-0.250 m/s²)(160 s)²
s = 1600 m - 204800 m
s = -203200 m (negative sign indicates the opposite direction)
Therefore, the puck will slide 203200 meters in the opposite direction.