a criket ball of mass 70g moving with velocity of 0.5m/s is stoppedby a player in 0.5s . what will be the force applied by the player to stop the ball
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Answer:
the mass of the ball is 70 g=.07kg
velocity of the ball is 0.5 m/s
time taken by the player is .05sec
from Newton 1st law
v=u+at
0.5=a×0.5
a=1m/sec
F=ma
F=.07×1
F=0.07N
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Mass = 70g
Initial velocity = 0.5 m/s
Final velocity = 0
Time = 0.5 s
Force = mass * acceleration
= m * a
= m ( v-u/t ). (a = v-u/t )
= mv-mu/t
= 70*0.5 - 70*0
= 35 - 0
= 35 N
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