A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and g = 9.8m / s2 , then the acceleration of the block
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0.88 m/s2
24-(0.4*9.8*5)=4.4
Acceleration= 4.4/5=0.88
Answered by
11
The answer to the given problem is as following
Given from the question,
Mass of block = 5kg
F = 24N
μ (friction) = 0.4 (force)
g (acceleration due to gravity)= 9.8 m/s²
friction, f = μmg = 0.4×5×9.8 = 19.6N
net force =F-f = 24-19.6 = 4.4N
acceleration = 4.4/5 = 0.88m/s²
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