Physics, asked by smanu1282, 1 year ago

A man of mass 70 kg stands on a weighing scale in a lift which is moving. What would be the readings on the scale in each of the following cases?a. upwards with a uniform speed of 10 m s⁻¹b. downwards with uniform acceleration of 5 m s⁻²c. upwards with a uniform acceleration of 5 m s⁻²d. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answers

Answered by Anonymous
15
(a) Mass of the man, m = 70 kg

Acceleration, a = 0

Using Newton’s second law of motion, we can write the equation of motion as:

R – mg = ma

Where, ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

∴R = mg

= 70 × 10 = 700 N

∴Reading on the weighing scale =700/g=700/10=70
(b) Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 downward

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma

R = m(g – a)

= 70 (10 – 5) = 70 × 5

= 350 N

∴Reading on the weighing scale =

350/g=350/10=35kg
(c) Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 upward

Using Newton’s second law of motion, we can write the equation of motion as:

R – mg = ma

R = m(g + a)

= 70 (10 + 5) = 70 × 15

= 1050 N

∴Reading on the weighing scale =

1050/10=105kg
(d) When the lift moves freely under gravity, acceleration a = g

Using Newton’s 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
I hope this will help you
Answered by gadakhsanket
7
Hey mate,

● Explained Answers-
# Given-
Mass of the man m = 70 kg

# Readings on weight scale when lift is moving -
(a) Upwards with a uniform speed of 10 m/s -
a = 0
By Newton’s second law of motion,
W – mg = ma
∴ W = mg
W = 70×10 = 700 N
∴ Apparent mass = W/g = 700/10 = 70 kg

(b) Downwards with a uniform acceleration of 5 m/s^2,
a = 5 m/s^2 downward
By Newton’s second law of motion,
W + mg = ma
W = m(g–a)
W = 70(10–5)
W = 350 N
∴ Apparent mass = W/g = 350/10 = 35 kg

(c) Upwards with a uniform acceleration of 5 m/s^–2 -
a = 5 m/s^2 upward
By Newton’s second law of motion,
W – mg = ma
W = m(g+a)
W = 70(10+5)
W = 1050 N
∴ Apparent mass = W/g = 1050/10 = 105 kg

(d) When the lift moves freely under gravity -
a = g
By Newton’s second law of motion,
W + mg = ma
W = m(g–a)
W = m(g–g) = 0
∴ Apparent mass = 0/g = 0 kg

Hope that was useful...
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