A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d
1 answer:
Answer:
x = 45 cm
Explanation:
Given that,
The length of a rod, L = 50 cm
Mass, m₁ = 0.2 kg
It is at 40cm from the left end of the rod.
We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.
The centre of mass of the rod is at 25 cm.
Taking moments of both masses such that,
The distance from the left end is 40+5 = 45 cm.
Hence, at a distance of 45 cm from the left end it will balance the rod.
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Answer:
Part a)
Part b)
if both sides are rough then it will reach the same height on the other side because the energy is being conserved.
Part c)
Since marble will go to same height when it is rough while when it is smooth then it will go to the height
so on smooth it will go to lower height
Explanation:
As we know by energy conservation the total energy at the bottom of the bowl is given as
here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy
now on the right side of the bowl there is no friction
so its rotational kinetic energy will not change and remains the same
so it will have
now we know that
so we have
so the height on the smooth side is given as
Part b)
if both sides are rough then it will reach the same height on the other side because the energy is being conserved.
Part c)
Since marble will go to same height when it is rough while when it is smooth then it will go to the height
so on smooth it will go to lower height
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
