Asked by Maria
A racket exerted an average force of 152.0 N on a ball initially at rest. If the ball has a mass of 0.070 kg and was in contact with the racket for 0.030 s, what was the kinetic energy of the ball as it left the racket?
Answers
Answered by
Henry
a = F/M = 152/0.07 = 2171 m/s^2
V = Vo + a*t
Vo = 0
a = 2171 m/s^2
t = 0.030 s.
V = ?
KE = o.5*M*V^2
V = Vo + a*t
Vo = 0
a = 2171 m/s^2
t = 0.030 s.
V = ?
KE = o.5*M*V^2
Answered by
Maria
Thank you so much Henry!
Answered by
Henry
Glad I could help!
Answered by
Jen
KE= 1/2 mv^2
= 1/2 (0.070 kg) v^2 ; Ft= mv
v= Ft/m
v=(152N)(0.030s)/0.070kg
V= 65 m/s
= 1/2 ((0.070 kg) (65 m/s) ^2
= 148.5 J
= 1/2 (0.070 kg) v^2 ; Ft= mv
v= Ft/m
v=(152N)(0.030s)/0.070kg
V= 65 m/s
= 1/2 ((0.070 kg) (65 m/s) ^2
= 148.5 J
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