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Hindi
4-2.Friction
medium

બ્લોક અને સપાટી વચ્ચેનો ઘર્ષણાંક $0.03$ છે. તો તંત્રનો પ્રવેગ ........ $m/s^{2}$ થશે . $(g = 10\,m{s^{ - 2}})$

A$1.8$
B$0.8$
C$1.4$
D$0.4$

Solution

$T – F = {m_2}a ⇒ T – \mu {m_2}g = {m_2}a$
$⇒ T – 0.03 \times 20 \times 10 = 20a$$ ⇒ T – 6 = 20a$…..(i)
${m_1}g – T = {m_1}a$
$⇒ 4 \times 10 – T = 4a$ $⇒ 40 – T = 4a$….(ii)
$a = 1.4m/{s^2}.$
Standard 11
Physics

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