Three masses M = 100kg, m₁ = 10kg and M₂ = 20kg are arranged in a system as shown in figure. All the surface are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m₂ moves upward with an acceleration of 2ms^-2. The value of F is:

(Take g = 10ms^-2)

Correct Option - 1


Detailed Solution:

Given Acceleration of m₂ is 2m/s²

(upward)

N = 20a              T – m₂g = m₂a

T – 20g = 20a

T = 200 + 20 × 2

T = 240 N

T = m₁ a₁

240 = 10a₁

a₁ = 24 m/s²

$\Rightarrow Ta+T{a}_1-T{a}_2=0$

$a_2=24+2=26m/s^2$

F – T – N = Ma₂

F – 240 = 100 × 26 + 20 × 26

F = 3360 N

<p>Given Acceleration of m₂ is 2m/s²</p><p> (upward)</p><p>N = 20a&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; T – m₂g = m₂a</p><p>T – 20g = 20a</p><p>T = 200 + 20 × 2</p><p>T = 240 N</p><p>T = m₁ a₁</p><p>240 = 10a₁</p><p>a₁ = 24 m/s²</p><p><strong><math><mrow><mo>⇒</mo><mi>T</mi><mi>a</mi><mo>+</mo><mi>T</mi><msub><mrow><mi>a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>−</mo><mi>T</mi><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mn>0</mn></mrow></math></strong></p><p><strong><math><mrow><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mn>2</mn><mn>4</mn><mo>+</mo><mn>2</mn><mo>=</mo><mn>2</mn><mn>6</mn><mi>m</mi><mo>/</mo><msup><mrow><mi>s</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></math></strong></p><p>F – T – N = Ma₂</p><p>F – 240 = 100 × 26 + 20 × 26</p><p>F = 3360 N</p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1754924987php1LSGZ3.jpeg" alt="" width="278" height="458" loading="lazy"></picture></div></div>

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