Answer:
0.38 kN
Explanation:
From newton's second law of motion,
F = m(v-u)/t........................ Equation 1
Where F = Average force exerted on the ball, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact.
Given: m = 55 g = 55/1000 = 0.055 kg, v = 102 ft/s = (102×0.3048) m/s = 31.0896 m/s, u = 0 m/s (initially at rest), t = 0.0045 s
Substitute into equation 1
F = 0.055(31.0896-0)/0.0045
F = 1.709928/0.0045
F = 379.984 N
F ≈ 0.38 kN
The average force exerted on the ball by the club is 0.38 kN. The force applied to an object is directly proportional to the product of mass and acceleration.
The force applied to an object is directly proportional to the product of mass and acceleration.
[tex]F = m\dfrac {(v-u)}t[/tex]
Where
[tex]F[/tex] - Average force exerted on the ball,
[tex]m[/tex] - mass of the ball = 55 g = 0.055 kg
[tex]v[/tex] - final velocity of the ball = 102 ft/s = 31.0896 m/s
[tex]u[/tex] - initial velocity of the ball = 0 m/s
[tex]t[/tex] - time of contact = 0.0045 s
Put the values in the formula,
[tex]F = 0.055\times \dfrac {(31.0896-0)}{0.0045}\\\\F = 379.984{\rm \ N}\\\\F = 0.38 \rm \ kN[/tex]
Therefore, the average force exerted on the ball by the club is 0.38 kN.
Learn more about Newton's second law of motion:
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