Respuesta :
Answer:
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, [tex]F_k[/tex], required is given by the formula, F = m × a,
From which we have;
[tex]F_k[/tex] = 0.5 × 0.125 = 0.0625 N
The kinetic frictional force required, [tex]F_k[/tex] = 0.0625 N
c. The coefficient of friction between the ice and puck, [tex]\mu_k[/tex], is given from the equation for the kinetic friction force as follows;
[tex]F_k = \mu_k \times Normal \ force \ of \ hockey \ puck = \mu_k \times 0.5 \times 9.8[/tex]
[tex]\mu_k = \dfrac{0.0625}{0.5 \times 9.8} \approx 0.012755[/tex]
The coefficient of friction between the ice and puck, [tex]\mu_k[/tex] ≈ 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 [tex]F_k = 0.012755 \times 1 \times 9.8 = 0.125 \ N[/tex]
The acceleration, a = [tex]F_k[/tex]/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, Â [tex]F_k = 0.125 \ N[/tex]
The coefficient of friction [tex]\mu_k[/tex] ≈ 0.012755 is constant
A) The Acceleration of the hockey puck = - 0.125 m/s²
B) The Kinetic frictional force needed is = Â 0.0625 N
C) Coefficient of friction between ice and puck ≈ 0.0128
D) Recalculated values for a 1 kg hockey puck
- Acceleration of hockey puck=  - 0.125 m/s² Â
- Kinetic frictional force = 0.125 N
- Coefficient of friction ≈ 0.0128
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
Fâ‚“ = 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.
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