A force of 19.6, N when applied parallel to t | vedclass

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(A) The limiting friction force $F_l$ is given by $F_l = \mu R = \mu mg$,where $\mu$ is the coefficient of static friction,$m$ is the mass,and $g$ is

Vedclass · Accepted Answer

$29.4$

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(A) The limiting friction force $F_l$ is given by $F_l = \mu R = \mu mg$,where $\mu$ is the coefficient of static friction,$m$ is the mass,and $g$ is the acceleration due to gravity.Since the surface remains the same,$\mu$ is constant. Thus,$F_l \propto m$.Given,for $m_1 = 10\, kg$,$F_{l1} = 19.6\, N$.When a $5\, kg$ mass is added,the new mass $m_2 = 10\, kg + 5\, kg = 15\, kg$.Using the proportionality $F_{l2} / F_{l1} = m_2 / m_1$,we get:$F_{l2} = F_{l1} 	imes (m_2 / m_1) = 19.6 	imes (15 / 10) = 19.6 	imes 1.5 = 29.4\, N$.

$A$ force of $19.6\, N$ when applied parallel to the surface just moves a body of mass $10\, kg$ kept on a horizontal surface. If a $5\, kg$ mass is kept on the first mass,the force applied parallel to the surface to just move the combined body is........ $N.$

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