A lift moves downwards with an acceleration of 9.8 m/sec2. the press exerted by a man on floor of lift is --
Answers
Answer:
(a). As the lift is moving at a uniform speed, acceleration a=0
R=mg
=70×10
⇒700N
Reading on the weighing scale
=700/g
=700/10=70kg
(b) Mass of the man, m=70kg
Acceleration, a=5m/s
2
downward
Using Newtons second law of motion, we can write the equation of motion as:
R+mg=ma
R=m(g−a)
=70(10−5)
=70×5
=350N
Reading on the weighing scale
⇒350g=350/10=35kg
(c) Mass of the man, m=70kg
Acceleration, a=5m/s
2
upward
Using Newtons second law of motion, we can write the equation of motion as:
R−mg=ma
R=m(g+a)
=70(10+5)=70×15
=1050N
Reading on the weighing scale
⇒1050/g=1050/10=105kg
(d) When the lift moves freely under gravity, acceleration a=g
Using Newtons second law of motion, we can write the equation of motion as:
R+mg=ma
R=m(g−a)
=m(g−g)=0
Reading on the weighing scale =0/g=0kg
The man will be in a state of weightlessness.
Explanation:
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