A hockey puck is hit on a frozen lake and starts moving with a speed of 13.80 m/s. Exactly 6.7 s later, its speed is 7.10 m/s.

a) what is the puck's average acceleration?

b) what is the coefficient of kinetic friction between the puck and the ice?

acceleration=changevelocity/time

forcefriction=mass*a you know a, and mass

force friction=mu*mg

mu*g=a
solve for mu.

To find the average acceleration of the hockey puck, we can use the following formula:

\[ \text{Average acceleration} = \frac{\text{change in velocity}}{\text{change in time}} \]

a) Change in velocity:
\[ \text{Change in velocity} = \text{final velocity} - \text{initial velocity} \]
\[ \text{Change in velocity} = 7.10 \, \text{m/s} - 13.80 \, \text{m/s} \]
\[ \text{Change in velocity} = -6.70 \, \text{m/s} \]

Change in time:
Given that the change in time is 6.7 s.

Now, we can calculate the average acceleration:
\[ \text{Average acceleration} = \frac{\text{change in velocity}}{\text{change in time}} \]
\[ \text{Average acceleration} = \frac{-6.70 \, \text{m/s}}{6.7 \, \text{s}} \]
\[ \text{Average acceleration} = -1.00 \, \text{m/s}^2 \]

Therefore, the puck's average acceleration is -1.00 m/s^2.

b) To determine the coefficient of kinetic friction between the puck and the ice, we can use the formula:

\[ F_{\text{friction}} = \mu \cdot F_{\text{norm}} \]

Where:
\[ F_{\text{friction}} \] is the force of friction,
\[ \mu \] is the coefficient of kinetic friction, and
\[ F_{\text{norm}} \] is the normal force.

When the puck is moving at a constant speed, the force of friction is equal and opposite to the applied force, since there is no net force. Therefore:

\[ F_{\text{friction}} = F_{\text{applied}} \]

Given that the mass of the puck is not provided, we can determine the force of friction using the equation of motion:

\[ F_{\text{applied}} = m \cdot a \]

Since the average acceleration is given (a = -1.00 m/s^2), we can rewrite the equation as:

\[ F_{\text{applied}} = m \cdot (-1.00) \]

Since the mass (m) cancels out, we can say:

\[ F_{\text{friction}} = F_{\text{applied}} \]

Therefore, the force of friction is equal to the applied force.

Hence, to calculate the coefficient of kinetic friction, we need the applied force and the normal force.

Could you please provide the applied force or any other missing information?

To find the answers to the given questions, we'll use the equations of motion.

a) To find the average acceleration of the puck, we'll use the equation:

average acceleration (a) = (final velocity - initial velocity) / time

Given:
Initial velocity (u) = 13.80 m/s
Final velocity (v) = 7.10 m/s
Time (t) = 6.7 s

Substituting the given values into our equation, we get:

a = (7.10 m/s - 13.80 m/s) / 6.7 s

Simplifying the equation yields:

a = (-6.70 m/s) / 6.7 s
a = -1.00 m/s^2

Therefore, the puck's average acceleration is -1.00 m/s^2. Note that the negative sign indicates that the acceleration is in the opposite direction to the initial velocity.

b) To find the coefficient of kinetic friction between the puck and the ice, we can use the equation of motion:

final velocity (v) = initial velocity (u) + (acceleration (a) * time (t))

Since the question states that the puck's speed is decreasing, we know that the acceleration is due to the friction force acting in the opposite direction of the initial velocity.

Considering the equation:

v = u + (a * t)

We can rearrange it to:

a = (v - u) / t

Substituting the given values:

a = (7.10 m/s - 13.80 m/s) / 6.7 s
a = -6.70 m/s / 6.7 s
a = -1.00 m/s^2

Since the acceleration is due to the friction force, we can equate it to the product of the coefficient of kinetic friction (μk) and the acceleration due to gravity (g), which is approximately 9.8 m/s^2:

-1.00 m/s^2 = μk * 9.8 m/s^2

Simplifying the equation gives:

μk = (-1.00 m/s^2) / 9.8 m/s^2
μk = -0.102

The coefficient of kinetic friction between the puck and the ice is approximately -0.102. Note that the negative sign indicates the direction of the friction force, opposite to the relative motion of the puck. The coefficient of friction is usually positive, so take the absolute value, which is 0.102.