A body of mass m and radius r is released from rest along a smooth inclined plane
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The force of acceleration on the plane of inclination θθ (which is the angle with the horizontal) will be
a=g×sin(θ)a=g×sin(θ)
(i.e when θ=0θ=0 plane is horizontal and there wont be any net force, and when θ=90∘θ=90∘ the net acceleration due to gravity is completely applied on the mass)
Now the inclined plane is of length hh
therefore final velocity v=2×a×h−−−−−−−−√v=2×a×h
substitution for a gives
v=2×g×sin(θ)×h−−−−−−−−−−−−−−√v=2×g×sin(θ)×h
where gg is the acceleration due to gravity.
.. hope this answer helps you
please mark it as brainliest
The force of acceleration on the plane of inclination θθ (which is the angle with the horizontal) will be
a=g×sin(θ)a=g×sin(θ)
(i.e when θ=0θ=0 plane is horizontal and there wont be any net force, and when θ=90∘θ=90∘ the net acceleration due to gravity is completely applied on the mass)
Now the inclined plane is of length hh
therefore final velocity v=2×a×h−−−−−−−−√v=2×a×h
substitution for a gives
v=2×g×sin(θ)×h−−−−−−−−−−−−−−√v=2×g×sin(θ)×h
where gg is the acceleration due to gravity.
.. hope this answer helps you
please mark it as brainliest
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