1) A 90 kg woman stands in an elevator. Find the force which the floor of the elevator exerts on the woman a) when the elevator has an upward acceleration of 2 m/sec2; b) when the elevator is rising at constant speed; c) when the elevator has a downward acceleration of 2 m/sec2.
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Let F be the force exerted by the floor of the lift on the woman.
The only two forces on the woman are weight and the Force F.
as per the Newtons law, the acceleration is equal to the net force on the object divided by its mass.
F - m g = m a
F = m (g + a) Here , a is positive upwards. F is positive upwards.
a) a = 2 m/sec/sec
F = 90 * ( 10 + 2) = 1080 N
b) a = 0
F = 90 * 10 = 900 N
c)
a = -2 m/sec/sec
F = 90 * ( 10 - 2) = 720 N
The only two forces on the woman are weight and the Force F.
as per the Newtons law, the acceleration is equal to the net force on the object divided by its mass.
F - m g = m a
F = m (g + a) Here , a is positive upwards. F is positive upwards.
a) a = 2 m/sec/sec
F = 90 * ( 10 + 2) = 1080 N
b) a = 0
F = 90 * 10 = 900 N
c)
a = -2 m/sec/sec
F = 90 * ( 10 - 2) = 720 N
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ANSWER:
Let F be the force exerted by the floor of the lift on the woman.
The only two forces on the woman are weight and the Force F.
as per the Newtons law, the acceleration is equal to the net force on the object divided by its mass.
F - m g = m a
F = m (g + a) Here , a is positive upwards. F is positive upwards.
a) a = 2 m/sec/sec
F = 90 * ( 10 + 2) = 1080 N
b) a = 0
F = 90 * 10 = 900 N
c)
a = -2 m/sec/sec
F = 90 * ( 10 - 2) = 720 N
Explanation:
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