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
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?
4 answers
Thank you so much Henry!
Glad I could help!
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