A puck of mass 0.03 kg slides across rough ice, experiencing a frictional force of 0.2 N. If it was moving at 36 m/s when I hit the ice patch,
A) How long did it take to stop?
B) What was the length of the ice patch?
I know I have initial velocity, mass and final velocity of 0. How do I find time? Is it considered uniform motion?
Since F=ma, the 0.2N friction causes a deceleration of .2/.03 = 1.67 m/s^2
so, its speed and position are described by
v = 36 - 1.67t
s = 36t - 0.83 t^2
You can find t when v=0 (stopped) and use that value to find the stopping position.
Not uniform motion, but uniform acceleration.
To find the time it takes for the puck to stop, we can use the equation of motion:
Final velocity (v) = Initial velocity (u) + (acceleration x time)
In this case, the acceleration is due to the frictional force, and since the puck is coming to a stop, the final velocity is 0 m/s.
A) How long did it take to stop?
We can rearrange the equation to solve for time:
0 = 36 m/s + (-0.2 N / 0.03 kg) x t
Simplifying,
0 = 36 m/s - (6.66 m/s^2) x t
Rearranging again,
(6.66 m/s^2) x t = 36 m/s
Dividing both sides by 6.66 m/s^2,
t = 36 m/s / 6.66 m/s^2
t ≈ 5.41 s
Therefore, it takes approximately 5.41 seconds for the puck to stop.
B) What was the length of the ice patch?
To find the length of the ice patch, we can use the equation of motion:
Final velocity^2 = Initial velocity^2 + 2 x acceleration x displacement
Since the final velocity is 0 m/s and the initial velocity is 36 m/s, the equation becomes:
0 = (36 m/s)^2 + 2 x (-0.2 N / 0.03 kg) x displacement
Simplifying,
0 = 1296 m^2/s^2 - (13.33 m/s^2) x displacement
Rearranging,
(13.33 m/s^2) x displacement = 1296 m^2/s^2
Dividing both sides by 13.33 m/s^2,
displacement = 1296 m^2/s^2 / 13.33 m/s^2
displacement ≈ 97.27 m
Therefore, the length of the ice patch is approximately 97.27 meters.
To find the time it takes for the puck to stop, you can use the equation of motion:
v = u + at
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
In this case, the initial velocity (u) is 36 m/s, the final velocity (v) is 0 m/s since the puck stops, and the acceleration (a) can be determined using Newton's second law:
F = ma
where F is the frictional force acting on the puck and m is the mass of the puck.
The equation can be rearranged to solve for acceleration (a):
a = F/m
Substituting the given values, we have:
a = 0.2 N / 0.03 kg
= 6.67 m/s²
Now, substituting the known values of u, v, and a into the equation of motion, we can solve for time (t):
0 = 36 m/s + (-6.67 m/s²) * t
Simplifying the equation:
-36 m/s = -6.67 m/s² * t
Dividing both sides by -6.67 m/s²:
t = 36 m/s / 6.67 m/s²
= 5.4 seconds
Therefore, it takes approximately 5.4 seconds for the puck to stop.
For part B) - determining the length of the ice patch:
You can use the equation of motion to determine the distance traveled by the puck during this time:
s = ut + 0.5at²
where:
s is the distance traveled,
u is the initial velocity,
t is the time, and
a is the acceleration (which we found to be -6.67 m/s²).
Since the puck starts at rest (u = 0), the equation simplifies to:
s = 0.5at²
Plugging in the values for a and t:
s = 0.5 * (-6.67 m/s²) * (5.4 seconds)²
Calculating the answer:
s ≈ 81.72 meters
Therefore, the length of the ice patch is approximately 81.72 meters.