formula for gravitational P.E =mgh
Solution:-mass=3kg height=5metre and gravity=9.8 or 10m/sec² so P.E=mgh , 3×9.8×5=147kgm²/sec²
Answer:
angular acceleration is -0.2063 rad/s²
Explanation:
Given data
mass m = 95.2 kg
radius r = 0.399 m
turning ω = 93 rpm
radial force N = 19.6 N
kinetic coefficient of friction μ = 0.2
to find out
angular acceleration
solution
we know frictional force that is = radial force × kinetic coefficient of friction
frictional force = 19.6 × 0.2
frictional force = 3.92 N
and
we know moment of inertia that is
γ = I ×α = frictional force × r
so
γ = 1/2 mr²α
α = -2f /mr
α = -2(3.92) /95.2 (0.399)
α = - 7.84 / 37.9848 = -0.2063
so angular acceleration is -0.2063 rad/s²
Answer:
Explanation:
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Answer:
Explanation:
Given
Force 
one at an angle of 
East of North and another at West of North
Net Force is in North Direction
Forces in horizontal direction will cancel out each other
thus Work done will be by north direction forces
here 
