Question 23 of 32
1 answer:
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Answer:
(a) Ff = 0.128 N
(b μk = 0.1135
Explanation:
kinematic analysis
Because the hockey puck moves with uniformly accelerated movement we apply the following formulas:
vf=v₀+a*t Formula (1)
d= v₀t+ (1/2)*a*t² Formula (2)
Where:
d:displacement in meters (m)
t : time in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s
Calculation of the acceleration of the hockey puck
We apply the Formula (1)
vf=v₀+a*t v₀=5.8 m/s , vf=0
0=5.8+a*t
-5.8 = a*t
a= -5.8/t Equation (1)
We replace a= -5.8/t in the Formula (2)
d= v₀*t+ (1/2)*a*t² , d=15.1 m , v₀=5.8 m/s
15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²
15.1= 5.8*t-2.9*t
15.1= 2.9*t
t = 15.1 / 2.9
t= 5.2 s
We replace t= 5.2 s in the equation (1)
a= -5.8/5.2
a= -1.115 m/s²
(a) Calculation of the frictional force (Ff)
We apply Newton's second law
∑F = m*a Formula (3)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Look at the free body diagram of the hockey puck in the attached graphic
∑Fx = m*a m= 115g * 10⁻³ Kg/g = 0.115g , a= -1.12 m/s²
-Ff = 0.115*(-1.115) We multiply by (-1 ) on both sides of the equation
Ff = 0.128 N
(b) Calculation of the coefficient of friction (μk)
N: Normal Force (N)
W=m*g= 0.115*9.8= 1.127 N : hockey puck Weight
g: acceleration due to gravity =9.8 m/s²
∑Fy = 0
N-W=0
N = W
N = 1.127 N
μk = Ff/N
μk = 0.128/1.127
μk = 0.1135
Answer:
the magnitude of the average force on the bumper is 3189.8 N
Explanation:
Given the data in the question;
In terms of force and displacement, work done is;
W = ×
W = ------- let this be equation 1
where F is force applied, x is displacement and θ is angle between force and displacement.
Now, since the displacement of the bumper and force acting on it is in the same direction,
hence, θ = 0°
we substitute into equation 1
W = 0°
W = ------- let this be equation 2
Now, using work energy theorem,
total work done on the system is equal to the change in kinetic energy of the system.
= ΔKE
= mv² -
mu² --------- let this be equation 3
where m is mass of object, v is final velocity, u is initial velocity.
from equation 2 and 3
=
mv² -
mu²
we make F, the subject of formula
F = ( v² - u² )
given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s
so we substitute
F = ( (0)² - (1.4)² )
F = 1627.45098 ( 0 - 1.96 )
F = 1627.45098 ( - 1.96 )
F = -3189.8 N
The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.
Therefore, the magnitude of the average force on the bumper is 3189.8 N