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Consider the Atwood Machine of the Previous Problem. the Larger Mass is Stopped for a Moment, 2.0 S After the System is Set into Motion. Find the Time that Elapses before the String is Tight Again. - Physics

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Sum

Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment, 2.0 s after the system is set into motion. Find the time that elapses before the string is tight again.

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Solution

a = 3.26 m/s2, T = 3.9 N
After 2 s, velocity of mass m1,
v = u + at = 0 + 3.26 × 2
= 6.52 m/s upward
At this time, m2 is moving 6.52 m/s downward.
At time 2 s, m2 stops for a moment. But m1 is moving upward with velocity 6.52 m/s. It will continue to move till final velocity (at highest point) becomes zero.

Here, v = 0, u = 6.52 m/s
a = −g = − 9.8 m/s2
v = u + at = 6.52 + (−9.8)t
$\Rightarrow t = \frac{6 . 52}{9 . 8} \approx \frac{2}{3} \sec$
After this time, the mass m1 also starts moving downward.
So, the string becomes tight again after $\frac{2}{3} s$

Concept: Newton’s Second 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 23 | Page 80
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