Question

Question: Find the time taken by the block to reach the bottom of inclined plane. E = 200 i
N/C, M = 1 kg, q = 5 mC, g = 10 m/s2
, µ = 0.2


Options:

(a) 1.35 s

(b) 1.65 s

(c) 1.9 s

(d) 2.3 s

#Refer the attachment!!​

Answer 1

Explanation:

${ \underline{ \underline{ \red{ \bf{Question:}}}}}$

• Find the time taken by the block to reach the bottom of inclined plane. E = 200 i N/C, M = 1 kg, q = 5 mC, g = 10 m/s2, µ = 0.2

• Options:
• (a) 1.35 s
• (b) 1.65 s
• (c) 1.9 s
• (d) 2.3 s

${ \underline{ \underline{ \bf{ \red{Answer:}}}}}$

$\bf { \underline{\pink{option(a) \: 1.35 \: s}}}$

${ \underline{ \underline{ \bf{ \red{Given :}}}}}$

• E = 200 i N/C,
• M = 1 kg,
• q = 5 mC,
• g = 10 m/s2
• µ = 0.2

${ \underline{ \underline{ \bf{ \red{To \: \: Find :}}}}}$

• The time taken by the block to reach the bottom of inclined plane.

${ \underline{ \underline{ \bf{ \red{Explanation :}}}}}$

Net Force among the incline .

$\huge{ \underline{ \boxed{ \rm{ \blue{F = mg \: sin \theta - ( \mu \: N \: + qE \: cos \theta)}}}}} \bigstar$

$\qquad \looparrowright \rm \: F = mg \: sin \theta \: - \mu(mg \: cos \theta + qE \: sin \theta) - qE \: cos \theta$

$\qquad \looparrowright \rm \: F =1 \times 10 \: sin \: 30 - 0.2(1 \times 10 \times cos \: 30 + 200 \times 5 \times {10}^{ - 3} \times sin30) - 200 \times 5 \times {10}^{ - 3} \: cos \: 30$

$\qquad \looparrowright \rm \: F =5 - 0.2(5 \sqrt{3} + 0.5) - \frac{ \sqrt{3} }{2}$

$\qquad \looparrowright \rm \: F =2.3 \: N$

$\longmapsto \bf \: a = F/m$

$\longmapsto \bf \: a = \frac{2.3}{1}$

$\longmapsto \bf \: a =2.3 \: \frac{m}{ {s}^{2} }$

Time taken to slide down 2m long

${ \boxed{ \tt{ \green{Incline \: t \: = \sqrt{ \frac{25}{a} } = \sqrt{ \frac{2 \times 2}{2.3} } = 1.32 \: s \: }}}}$

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Hope it will help you S...

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Answer 2

ANSWER:

• Time taken by the block to reach the bottom of inclined plane = 1.35 s

GIVEN:

• E = 200 N/C.

• M = 1 kg.

• q = 5 mC.

• g = 10 m/s².

• µ = 0.2.

TO FIND:

• Time taken by the block to reach the bottom of inclined plane.

EXPLANATION:

$\sf From \ the \ diagram, \\ \sf ma = mg sin 30^{\circ} -\mu R - qE cos30^{\circ} \\ \\ \\ \sf a = \dfrac{10}{2} - 0.2 \left(mgcos 30^{\circ} + qE sin30^{\circ}\right) - q E cos30^{\circ} \\ \\ \\ \sf a = 5 - 0.2 \left(10 \times \dfrac{ \sqrt{3} }{2}+ \dfrac{5 \times {10}^{ - 3} \times 200}{2}\right) - \dfrac{5 \times {10}^{ - 3} \times 200 \times \sqrt{3} }{2} \\ \\ \\ \sf a = 5 - 0.2 (5 \sqrt{3} + 0.5) + \dfrac{1.732} {2} \\ \\ \\ \sf a = 5- \sqrt{3} - 0.1 - 0.866 \\ \\ \\ \sf a = 2.302 \ m {s}^{ - 2} \\ \\ \\ \sf From \ Pythagoras \ theorem, \\ \sf Length \ of \ inclined \ path = \dfrac{1}{sin {30}^{ \circ} } = 2 \ m. \\ \\ \\\blue{\boxed{ \bold{ \green{ \large{s = \dfrac{1}{2} a {t}^{2}}}}}} \\ \\ \\ \sf 2 = \dfrac{1}{2} \times 2.302 \times {t}^{2} \\ \\ \\ \sf {t}^{2} = \dfrac{4}{2.302} \\ \\ \\ \sf t = 1.32 \ s \approx 1. 35 \ s$

Hence the time taken by the block to reach the bottom of inclined plane = 1.35 s.