A tennis racket exerts 150.0 N on the ball. If the ball has a mass of .6 kg and is in contact with the strings of the racket for .03 seconds, what is the kinetic energy of the ball as it leaves the racket?
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Question is incomplete. I’m assuming the ball to be initially at rest.
a = F / m
= 150 N / (0.6 kg)
= 250 m/s²
v = u + at
= 0 + (250 m/s² × 0.03 s)
= 7.5 m/s
K.E = 0.5mv²
= 0.5 × 0.6 kg × (7.5 m/s)²
= 16.875 Joules
Kinetic energy of the ball as it leaves the racket is 16.875 Joules
a = F / m
= 150 N / (0.6 kg)
= 250 m/s²
v = u + at
= 0 + (250 m/s² × 0.03 s)
= 7.5 m/s
K.E = 0.5mv²
= 0.5 × 0.6 kg × (7.5 m/s)²
= 16.875 Joules
Kinetic energy of the ball as it leaves the racket is 16.875 Joules
Anonymous:
Superb ! Keep going
Answered by
3
Solution :-
Using, impulse formula, get
velocity v = (150 * 0.03)/0.6 = 7.5 m/s
Then, using kinetic energy formula
KE = 0.5 * 0.6 * (7.5)²
= 16.875 J
Hopes this helps you !
Using, impulse formula, get
velocity v = (150 * 0.03)/0.6 = 7.5 m/s
Then, using kinetic energy formula
KE = 0.5 * 0.6 * (7.5)²
= 16.875 J
Hopes this helps you !
Excellent answer ! BHAI
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