a 160g hockey puck slides horizontally along the ice, and the coefficient of kinetic friction between puck and ice is 0.15. if the puck starts sliding at 3 m/s before it comes to a stop it travels 3.6m how do you get to this answer?
a 160 g hockey puck slides horizontally along the ice, and the coefficient of kinetic friction between puck and ice is 0.15. if the puck a tarts sliding at 3 m/s, how far will it travel before it comes to a stop?
To find the answer, we can use the following steps:
Step 1: Identify the given information:
- Mass of the hockey puck (m) = 160 g = 0.16 kg
- Coefficient of kinetic friction (μ) = 0.15
- Initial velocity (u) = 3 m/s
- Distance traveled (s) = 3.6 m
Step 2: Calculate the frictional force:
The frictional force (Ff) can be calculated using the equation:
Ff = μ * (mass of the puck) * (acceleration due to gravity)
The mass of the puck needs to be converted to kilograms:
Mass of the puck (m) = 0.16 kg
The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Substituting the values into the equation, we have:
Ff = 0.15 * 0.16 kg * 9.8 m/s^2
Step 3: Calculate the force needed to bring the puck to a stop:
The force needed to bring the puck to a stop (Fnet) can be calculated using the equation:
Fnet = mass of the puck * deceleration
Since the puck comes to a stop, its final velocity (v) is zero, and the deceleration (a) can be calculated using the equation:
v^2 = u^2 + 2 * a * s
Substituting the given values, we have:
0 = (3 m/s)^2 + 2 * a * 3.6 m
Simplifying the equation:
0 = 9 m^2/s^2 + 7.2 a
Rearranging the equation to solve for a, we have:
a = -9 m^2/s^2 / 7.2
a = -1.25 m/s^2
Substituting the calculated deceleration (a) and mass of the puck (m) into the equation for the force (Fnet), we have:
Fnet = 0.16 kg * (-1.25 m/s^2)
Step 4: Relating the force and the frictional force:
The force needed to bring the puck to a stop (Fnet) is equal to the frictional force (Ff). Therefore:
Ff = Fnet
Step 5: Solving for distance traveled:
The work done by the frictional force is given by:
Work = Force * Distance
The work done by friction (Wf) is equal to the force of friction (Ff) multiplied by the distance traveled (s). Therefore:
Wf = Ff * s
However, the work done by friction is given by:
Wf = μ * (mass of the puck) * (acceleration due to gravity) * s
Setting the two equations equal to each other, we have:
μ * (mass of the puck) * (acceleration due to gravity) * s = Fnet * s
Substituting the given and calculated values:
0.15 * 0.16 kg * 9.8 m/s^2 * 3.6 m = 0.16 kg * (-1.25 m/s^2) * 3.6 m
Simplifying the equation:
9.828 N * m = -1.152 N * m
Step 6: Calculate the distance traveled (s):
Dividing both sides of the equation by the force of friction (Ff):
9.828 N * m / 0.15 * 0.16 kg * 9.8 m/s^2 = -1.152 N * m / force of friction
Simplifying the equation:
s = -1.152 N * m / 0.15 * 0.16 kg * 9.8 m/s^2
Calculating the distance traveled (s):
s ≈ 3.593 m
Therefore, the approximate distance traveled by the hockey puck is 3.6 m.
To find the answer, we need to analyze the forces acting on the hockey puck and use Newton's laws of motion. In this case, the force of kinetic friction is the main force opposing the motion of the puck.
The formula for kinetic friction is given by:
Frictional force = coefficient of kinetic friction * normal force
The normal force is the force exerted by the ice on the puck, and it is equal to the weight of the puck, which can be calculated by:
Weight = mass * acceleration due to gravity
In this case, the mass of the hockey puck is given as 160 grams or 0.16 kg, and the acceleration due to gravity is approximately 9.8 m/s².
Weight = 0.16 kg * 9.8 m/s² = 1.568 N
Now, we can find the frictional force:
Frictional force = 0.15 * 1.568 N = 0.2352 N
The frictional force acts in the opposite direction to the motion of the puck, so we can define the net force as:
Net force = Frictional force
Using Newton's second law of motion, we know that the net force is equal to the mass of the object multiplied by its acceleration:
Net force = mass * acceleration
For this case, the puck comes to a stop, so the acceleration is negative to indicate the direction opposing its motion.
Net force = 0.16 kg * acceleration
Therefore:
-0.2352 N = 0.16 kg * acceleration
Solving for the acceleration:
acceleration = (-0.2352 N) / (0.16 kg) = -1.47 m/s²
This negative acceleration tells us that the puck is decelerating or slowing down.
We know the initial velocity is 3 m/s, the acceleration is -1.47 m/s², and we want to find the distance or displacement traveled.
The equation for constant acceleration is:
final velocity² = initial velocity² + 2 * acceleration * distance
Since the final velocity is 0 (the puck comes to a stop), we can solve the equation for distance:
(0)² = (3)² + 2 * (-1.47) * distance
0 = 9 - 2.94 * distance
Rearranging the equation:
2.94 * distance = 9
distance = 9 / 2.94 = 3.06 m
Therefore, the puck travels approximately 3.06 meters before it comes to a stop.
normal force = weight = .160*9.8
friction force = -.15*.160*9.8
a = F/m = -.15*.160*9.8/.160
=-.15*9.8 =-1.47
0 = vi + at = 3 - 1.47 t
so
t to stop = 3/1.47 = 2.04 seconds to stop
d = vi t - .5*1.47 t^2
=3*2.04 - .735*2.04^2
= 6.02-3.06
= 3 meters