I’M DESPERATE PLZ HELP 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?
b. What Kinetic Frictional Force is needed?
c. What is the co-efficient of friction between the ice and puck?
d. Recalculate a, b &c for a 1 kg hockey puck. (Does the weight change a or c?)

A) The Acceleration of the hockey puck = - 0.125 m/s²

B) The Kinetic frictional force needed is =  0.0625 N

C) Coefficient of friction between ice and puck ≈ 0.0128

D) Recalculated values for a 1 kg hockey puck

Given data :

Initial Mass of hockey puck = 0.5 kg

Initial velocity ( u ) = 5 m/s

Distance the puck slows down ( s )= 100 meters

New mass of hockey puck = 1 kg

Final velocity ( v ) = 0 m/s

A) Determine the acceleration of the hockey puck.

Applying the kinematics equation

V² = u² + 2as ( 1 )

∴ a = V² - u² / ( 2 * s )

     = ( 0 - 5² ) / ( 2 * 100 )

     = -25 / 200 = -0.125 m/s²

B) Determine the kinetic frictional force needed

Fₓ = m * a

  = 0.5 * ( 0.125 )

  = 0.0625

C) Determine the coefficient of friction between the ice and puck

Applying the kinematic equation

Fₓ = μ * Force on hockey

∴ μ = ( 0.0625 ) / ( 0.5 * 9.8 )

      = 0.0128

D) Given that coefficient of friction = constant

a) Acceleration of hockey = - 0.125 m/s²

b) Kinetic frictional force = 0.0128 * 1 * 9.81 = 0.125 N

c) coefficient of friction = constant  = 0.0128

note : The change in weight does not affect  acceleration and coefficient of friction.

Hence we can conclude that The answers to your questions are as listed above.

Learn more : link

a. The acceleration of the hockey puck is -0.125 m/s².

b. The kinetic frictional force needed is 0.0625 N

c. The coefficient of friction between the ice and puck, is approximately 0.012755

d. The acceleration is -0.125 m/s²

The frictional force is 0.125 N

The coefficient of friction is approximately 0.012755

Explanation:

a. The given parameters are;

The mass of the hockey puck, m = 0.5 kg

The starting velocity of the hockey puck, u = 5 m/s

The distance the puck slides and slows for, s = 100 meters

The acceleration of the hockey as it slides and slows and stops, a = Constant acceleration

The velocity of the hockey puck after motion, v = 0 m/s

The acceleration of the hockey puck is obtained from the kinematic equation of motion as follows;

v² = u² + 2·a·s

Therefore, by substituting the known values, we have;

0² = 5² + 2 × a × 100

-(5²) = 2 × a × 100

-25 = 200·a

a = -25/200 = -0.125

The constant acceleration of the hockey puck, a = -0.125 m/s².

b. The kinetic frictional force, , required is given by the formula, F = m × a,

From which we have;

= 0.5 × 0.125 = 0.0625 N

The kinetic frictional force required, = 0.0625 N

c. The coefficient of friction between the ice and puck, , is given from the equation for the kinetic friction force as follows;

The coefficient of friction between the ice and puck, ≈ 0.012755

d. When the mass of the hockey puck is 1 kg, we have;

Given that the coefficient of friction is constant, we have;

The frictional force

The acceleration, a = /m = 0.125/1 = 0.125 m/s², therefore, the magnitude of the acceleration remains the same and given that the hockey puck slows, the acceleration is -0.125 m/s² as in part a

The frictional force as calculated here,  

The coefficient of friction ≈ 0.012755 is constant

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.