Answer: 2.63 m/s
Explanation:
The question in english is written below:
During an attemp of croquet game, a 0.52 kg ball at rest on the grass is hit by a mallet with an average force of 190 N, if the mallet is in contact with the ball for [tex]7.2(10)^{-3} s[/tex], Whay is the speed of the ball just after being hit?
According to Newton's second law of motion, the force [tex]F[/tex] exerted on an object is directly proportional to its mass [tex]m[/tex] and acceleration [tex]a[/tex]:
[tex]F=m.a[/tex] (1)
On the other hand, acceleration is defined as the variation of velocity [tex]V[/tex] in time [tex]t[/tex]:
[tex]a=\frac{V}{t}[/tex] (2)
Substituting (2) in (1):
[tex]F=m\frac{V}{t}[/tex] (3)
Where:
[tex]F=190 N[/tex]
[tex]m=0.52 kg[/tex]
[tex]t=7.2(10)^{-3} s[/tex]
Isolating the velocity [tex]V[/tex] from (3):
[tex]V=\frac{Ft}{m}[/tex] (4)
[tex]V=\frac{(190 N)(7.2(10)^{-3} s)}{0.52 kg}[/tex]
Finally:
[tex]V=2.63 m/s[/tex]