Question

A 0.5 kg hockey puck slides across the ice starting with a velocity of 5 m/s. The puck slides and slows for 100 meters under constant acceleration and then stops.

a. What is the acceleration of the hockey puck? (0.125 m/s²)
b. What Kinetic Frictional Force is needed? (0.0625 N)
c. What is the coefficient of friction between the ice and puck? (0.013)
d. Recalculate a, b, & c for a 1 kg hockey puck. (Does the weight change a or c?)

Answer 1

The acceleration of the hockey puck is -0.125 m/s². The Kinetic Frictional Force needed is -0.0625 N. The coefficient of friction between the ice and puck is 0.013.

Let's go step by step:

a. Acceleration (a):

We can use the kinematic equation: v^2 = u^2 + 2as. Since v = 0, the equation simplifies to: 0 = 5^2 + 2a * 100. Solving for a: a = -5^2 / (2 * 100) = -0.125 m/s^2. So, the correct answer for part a is -0.125 m/s^2.

b. Kinetic Frictional Force (F_friction):

We can use Newton's second law: F_net = ma. The net force is the force of friction (F_friction), and we can express it as: F_friction = m * a. Substitute the values: F_friction = 0.5 kg * (-0.125 m/s^2) = -0.0625 N. So, the correct answer for part b is -0.0625 N.

c. Coefficient of Friction (μ):

The frictional force can also be expressed as F_friction = μ * N, where N is the normal force. Since the puck is on a horizontal surface, the normal force is equal to the weight of the puck (N = mg). So, μ * mg = -0.0625 N. Solve for μ: μ = -0.0625 / (0.5 * 9.8) ≈ 0.013. So, the correct answer for part c is 0.013.

d. Recalculate for a 1 kg hockey puck:

For a 1 kg puck, the mass (m) changes to 1 kg.

i. Acceleration (a):

Using the same kinematic equation: 0 = 5^2 + 2a * 100. Solving for a: a = -5^2 / (2 * 100) = -0.125 m/s^2. The acceleration is the same for a 1 kg puck.

ii. Kinetic Frictional Force (F_friction):

F_friction = 1 kg * (-0.125 m/s^2) = -0.125 N. The frictional force increases to -0.125 N for a 1 kg puck.

iii. Coefficient of Friction (μ):

μ = -0.125 / (1 * 9.8) ≈ 0.013. The coefficient of friction remains the same at 0.013.

Changing the mass of the puck to 1 kg does affect the kinetic frictional force, but does not affect the acceleration or the coefficient of friction.

Answer 2

The acceleration of the hockey puck is -0.125 m/s². The kinetic frictional force needed is 0.0625 N. The coefficient of friction between the ice and puck is 0.013. The weight does not change the acceleration or coefficient of friction.

Step-by-step explanation:

To find the acceleration of the hockey puck, we can use the equation of motion:

vf^2 = vi^2 + 2ad

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance.

Since the puck starts with a velocity of 5 m/s and comes to a stop, vf = 0. The distance traveled is 100 meters. Plugging the values into the equation, we get:

0 = 5^2 + 2a(100)

Simplifying, we find that a = -0.125 m/s^2.

To find the kinetic frictional force, we can use the equation:

Ff = μN

where Ff is the frictional force, μ is the coefficient of friction, and N is the normal force. Since the puck is moving horizontally, there is no vertical force, so N = mg, where m is the mass of the puck and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we get:

Ff = (0.013)(0.5 kg)(9.8 m/s^2) = 0.0625 N.

The acceleration and coefficient of friction do not depend on the weight of the hockey puck, so they remain the same for a 1 kg hockey puck.