Laws of Motion

JEE Main level, single correct, medium

138 of 210

Question 138 of 210

Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.040.04, the acceleration of the system in ms−2m{s}^{-2} is: (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless) :

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Answer: (C) 22


Maximum kinetic friction acting,
fk=μN=0.04×20g=8{f}_{k}=\mu N=0.04\times 20g=8 Newton
Therefore, acceleration of the system will be
a=60−826=2ms−2a=\dfrac{60-8}{26}=2m{s}^{-2}

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