A bullet of mass 0.4 kg with velocity 90 m/s enters a heavy wooden block and is stopped after distance of 60cm. what is the average resistive force exeted by block on the bullet
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DISTANCE = 60 CM = 0.6 M
MASS = 0.4 KG
INITIAL VELOCITY = 90 M/S
FINAL VELOCITY = 0 M/S
FIRST OF ALL,
FINDING ACCELERATION, USING THE THIRD EQUATION OF UNIFORMLY ACCELERATED MOTION,
V² = U² + 2AS
0² = 90² + 2×A×0.6
0 = 8100 + 1.2A
1.2A = -8100
A = -8100/1.2
A = -6750 M/S²
ACCORDING TO NEWTON'S SECOND LAW OF MOTION,
WE KNOW,
FORCE = CHANGE IN MOMENTUM / TIME TAKEN
FORCE = 0.4 × -6750
FORCE = -2700 NEWTONS
PLZ MARK IT AS BRAINLIEST ANSWER AND DROP A ♥
MASS = 0.4 KG
INITIAL VELOCITY = 90 M/S
FINAL VELOCITY = 0 M/S
FIRST OF ALL,
FINDING ACCELERATION, USING THE THIRD EQUATION OF UNIFORMLY ACCELERATED MOTION,
V² = U² + 2AS
0² = 90² + 2×A×0.6
0 = 8100 + 1.2A
1.2A = -8100
A = -8100/1.2
A = -6750 M/S²
ACCORDING TO NEWTON'S SECOND LAW OF MOTION,
WE KNOW,
FORCE = CHANGE IN MOMENTUM / TIME TAKEN
FORCE = 0.4 × -6750
FORCE = -2700 NEWTONS
PLZ MARK IT AS BRAINLIEST ANSWER AND DROP A ♥
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