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The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.
<h3>What is friction force?</h3>
The friction force is the opposing force which acts on the object which is in relative motion.
The driving force is equal and opposite to the friction force acting between road and bicycle.
Friction force = 400N
The friction force between rider and bike is zero.
So the magnitude of sum of friction force = 400N +0 = 400N
Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.
Learn more about friction force.
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KE = (1/2)·(mass)·(speed)²
KE = (1/2)·(50 kg)·(18 m/s)²
KE = (25 kg)·(324 m²/s²)
KE = 8,100 kg-m²/s²
KE = 8,100 Joules
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
