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⏺ A block of mass 5 kg on a rough horizontal surface and is rest . Now a force of 24 N is imparted to it with negligble impulse . If the coefficient of kineitic friction is 0.4 and g=9.8 m/s^2 , then calculate the acceleration of the block.
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Answer:
a=0.88 m/s²
Explanation:
we know that formula to be applied is
F-Ff=ma
where
F=force applied
Ff=frictional force
a=acceleration produced
m=mass of the object
We subtract the force Ff because Ff acts in opposite direction, it tries to stop the block and opposes relative motion.
F=24N
m=5kg
Ff=μN
N=mg
where N is the normal reaction
Ff=0.4*5*9.8 N
=19.6 N
plugging in the values
F-Ff=ma
a=(F-Ff)/m
=(24-19.6)/5
=0.88 m/s²
Hope this helps
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