Answer:
I was going to give you the paper where I saw it but since you are not giving enough points I can not give you so I am only going to give you some of these that are here sorry
Explanation:
1.
x=5
7.
5,12,13
9.
When Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will _____.
2 answers:
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Answer:
a) 567J
b) 283.5J
c)850.5J
Explanation:
The expression for the translational kinetic energy is,
Substitute,
14kg for m
9m/s for v
The translational kinetic energy of the center of mass is 567J
(B)
The expression for the rotational kinetic energy is,
The expression for the moment of inertia of the cylinder is,
The expression for angular velocity is,
substitute
1/2mr² for I
and vr for w
in equation for rotational kinetic energy as follows:
The rotational kinetic energy of the center of mass is 283.5J
(c)
The expression for the total energy is,
substitute 567J for E(r) and 283.5J for E(R)
The total energy of the cylinder is 850.5J
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
