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When the Elevator Goes up with Acceleration 1.2 M/S2: When the Elevator Goes up with Acceleration 1.2 M/S2: - Physics

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Sum

A pendulum bob of mass 50 g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goes up with acceleration 1.2 m/s2, (b) goes up with deceleration 1.2 m/s2, (c) goes up with uniform velocity, (d) goes down with acceleration 1.2 m/s2, (e) goes down with deceleration 1.2 m/s2 and (f) goes down with uniform velocity.

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Solution

(a) When the elevator goes up with acceleration 1.2 m/s2:

\[T = mg + ma\]
⇒ T = 0.05 (9.8 + 1.2) = 0.55 N
(b) Goes up with deceleration 1.2 m/s2 :

\[T = mg + m\left( - a \right) = m\left( g - a \right)\]
⇒ T  = 0.05 (9.8 − 1.2) = 0.43 N

(c) Goes up with uniform velocity:

\[T = mg\]
⇒ T = 0.05 × 9.8 = 0.49 N

(d) Goes down with acceleration 1.2 m/s2 :

\[T + ma = mg\]
\[ \Rightarrow T = m\left( g - a \right)\]
⇒ T = 0.05 (9.8 − 1.2) = 0.43 N

(e) Goes down with deceleration 1.2 m/s2 :

\[T + m\left( - a \right) = mg\]
\[ \Rightarrow T = m\left( g + a \right)\]
⇒ = 0.05 (9.8 + 1.2) = 0.55 N

(f) Goes down with uniform velocity:

\[T = mg\]
 T = 0.05 × 9.8 = 0.49 N

Concept: Newton's Third Law of Motion
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 5 Newton's Laws of Motion
Q 14 | Page 80
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