Answer:
Explanation:
The work done by the force to stop the ball will decrease the kinetic energy of it. so we can write
W = - ΔK. E
F.d = - ([tex]\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}[/tex])
As the final velocity is zero.
F = [tex]\frac{mv^{2} }{2d}[/tex]
The horizontal force exerted on the ball by the softball player's hand is [tex]\bold{F = \dfrac{m v_i}{2d}}[/tex].
Given to us,
mass of ball = m,
horizontal speed of ball = [tex]\bold{v_i}[/tex],
softball player's hand displacement = d,
Work done by a system is always given as the product of force and displacement.
[tex]\bold{Work\ done = Force\times displacement = F\times a}[/tex]
Kinetic energy is the energy that an object possesses due to its motion.
[tex]\bold{KE = \dfrac{1}{2} \times mass\times velocity = \dfrac{1}{2} \times m\times v}[/tex]
The work done by the force to stop the ball will decrease its kinetic energy. therefore, according to the law of conservation of energy,
[tex]\bold{Work\ done= -(change\ in\ kinetic\ energy)}[/tex]
[tex]F \times d =-[(\frac{1}{2} \times m\times v_f)-(\frac{1}{2} \times m\times v_i)][/tex]
As [tex]\bold{v_f}[/tex] = 0,
[tex]F \times d =-[(\frac{1}{2} \times m\times v_f)-(\frac{1}{2} \times m\times v_i)]\\\\F \times d =-[(\frac{1}{2} \times m\times 0)-(\frac{1}{2} \times m\times v_i)]\\\\F \times d =-[-(\frac{1}{2} \times m\times v_i)]\\\\F \times d =(\frac{1}{2} \times m\times v_i)\\\\F = \dfrac{m\times v_i}{2 \times d}[/tex]
Hence, the horizontal force exerted on the ball by the softball player's hand is [tex]\bold{F = \dfrac{m v_i}{2d}}[/tex].
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