Find the acceleration of 2 kg block shown in the diagram. (neglect friction)
Find the acceleration of 2 kg block shown in the diagram. (neglect friction)
Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>4</mn> <mi>g</mi> </mrow> <mrow> <mn>1</mn> <mn>5</mn> </mrow> </mfrac> </mrow> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mi>g</mi> </mrow> <mrow> <mn>1</mn> <mn>5</mn> </mrow> </mfrac> </mrow> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mi>g</mi> </mrow> <mrow> <mn>1</mn> <mn>5</mn> </mrow> </mfrac> </mrow> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mi>g</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> </span></p>
12 Views|Posted 4 months ago
Asked by Shiksha User
1 Answer
A
Answered by
4 months ago
Correct Option - 1
Detailed Solution:
For 2 kg block
T – 2g sin37 = 2a . (i)
For 4 kg block
4g – 2T =
2g – T = a . (ii)
T = (2g – a)
2g – a – 2g * = 2a
3a = 2g *
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= 0.92 * 1260 = 1161 m/s
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