Answer:
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Answer:
v = 4.4271 m/s
Explanation:
Given
m = 3 Kg
R = 0.2 m
∅ = 15°
h = 1.5 m
g = 9.8 m/s²
v = ?
Ignoring frictional losses, at the bottom of the plane
Total kinetic energy is = Potential Energy at the top of plane
Using Law of conservation of energy we have
U = Kt + Kr
m*g*h = 0.5*m*v² + 0.5*I*ω²
knowing that
Icylinder = 0.5*m*R²
ω = v/R
we have
m*g*h = 0.5*m*v² + 0.5*(0.5*mR²)*(v/R)² = 0.75*m*v²
⇒ v = √(g*h/0.75) = √(9.8 m/s²*1.5 m/0.75)
⇒ v = 4.4271 m/s
The net torque exerted by the children on the branch of the tree is 1382 N-m.
The torque exert by the kids is calculated as
T= 45.6*9.8*1.28*cos27.5°+36*9.8*(2.25-1.28)cos27.5°
T=1382 N-m
Hence, The net torque exerted by the children on the branch of the tree is 1382 N-m.
