Physics, 29.07.2019 17:30 duplessistoccara

Ahockey puck is hit on a frozen lake and starts moving with a speed of 12.0 m/s. exactly 5.0 s later, its speed is 6.0 m/s. what is the puck's average acceleration? what is the coefficient of kinetic friction between the puck and the ice?

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Answer from: mrmendrala

a. The puck's average acceleration is equal to -1.2 m/s^2 .

b. The coefficient of kinetic friction between the puck and the ice is 0.122.

Given the following data:

Initial speed = 12.0 m/sTime = 5.0 seconds.Final speed = 6.0 m/s

a. To find the puck's average acceleration:

Mathematically, acceleration is given by the formula;

$Acceleration = \frac{V_f\; - \;V_i}{t}$

Where:

V is the final velocity. U is the initial velocity. t is the time measured in seconds.

$Acceleration = \frac{6\; - \;12}{5}\\\\Acceleration = \frac{-6}{5}$

Average acceleration = -1.2 m/s^2

b. To find the coefficient of kinetic friction between the puck and the ice:

The frictional force acting between the puck and the ice is given by the formula;

$F = uN = -umg\\\\F = -umg\\\\\frac{F}{m} = -ug$ ....equation 1.

But, $a = \frac{F}{m}$ ....equation 2.

We know that acceleration due to gravity (g) of an object is equal to 9.8 meter per seconds square.

$a = -ug\\\\u = \frac{-a}{g}\\\\u = \frac{-(-1.2)}{9.8} \\\\u = \frac{1.2}{9.8}$

Coefficient of kinetic friction, u = 0.122

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Answer from: nikki8240

1) In an uniformly accelerated motion, the acceleration is given by:
$a= \frac{v_f-v_i}{\Delta t}$
where v_f

is the final speed, v_i

is the initial speed, and $\Delta t$ is the time interval between the initial and final point of the motion.

Using the data of the problem: v_i = 12.0 m/s

, and the time $\Delta t = 5.0 s$ , the acceleration is
$a= \frac{6 m/s-12 m/s}{5.0 s}=-1.2 m/s^2$
where the negative sign means that the hockey puck is decelerating.

2) The frictional force F between the puck and ice is responsible for the deceleration of the puck, and for second Newton's law this force is equal to the product between the mass of the puck m and the acceleration:
F=ma

(1)
The frictional force can also be written as
$F=-\mu_D mg$ (2)
where $\mu_D$ is the coefficient of kinetic friction, and where the negative sign is due to the fact that the frictional force acts against the direction of motion of the puck. By equilizing (1) and (2) we can find the value of this coefficient:
$ma=-\mu_D mg$
$\mu_D = -\frac{a}{g}= -\frac{-1.2 m/s^2}{9.81 m/s^2}=0.12$

Answer from: Quest

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

Answer from: Quest

The center is basically the heart of it

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Ahockey puck is hit on a frozen lake and starts moving with a speed of 12.0 m/s. exactly 5.0 s later...

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