A body of mass 2kg is released from a point 100m above the ground. Calculate its kinetic energy 80m from the point of release
Kinetic energy (K.E) = 1/2 mv2 ------------ eqn(1)
Where m is the mass and v is the velocity
from newton's third law of motion, v2 = u2 + 2as,
If the body/object is falling under gravity, the relation becomes:
v2 = u2 + 2gh ------------- eqn(2)
Where g, acceleration due to gravity = 10m/s2 and h is the height.
Since the object (released at an initial height of 100m) falls 80m (free falling under gravity), eqn(2) can be used to find the velocity at which the body is falling.
v2 = u2 + 2(10 × 80)
Also u = 0
v2 = 02 + 1600
v2 = 1600
v = 40m/s
From eqn(1), K.E = 1/2 × 2 × 402
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