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3 years ago

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A 70 kg base runner begins his slide into second base while moving at a speed of 4.35 m/s. He slides so that his speed is zero j

ust as he reaches the base. The acceleration of gravity is 9.8 m/s 2.
A. What is the magnitude of the mechanical energy lost due to friction acting on the runner?
B. How far does he slide?

1 answer:

sladkih [1.3K]3 years ago

3 0

Answer

given,

Mass of the runner, M = 70 Kg

speed of the runner on the second base = 4.35 m/s

speed at the base = 0 m/s

Acceleration due to gravity,g = 9.8 m/s²

a) magnitude of mechanical energy lost

Mechanical energy lost is equal top gain in kinetic energy

$ME_{Lost}=\dfrac{1}{2}mv^2$

$ME_{Lost}=\dfrac{1}{2}\times 70\times 4.35^2$

$ME_{Lost}=662.29\ J$

b) Work done = Force x displacement

W = F. x

F = μ mg

W = μ mg . x

Work done is equal to 662.29 J

$x=\dfrac{W}{\mu m g}$

using the coefficient of the friction,μ = 0.7

$x=\dfrac{662.29}{ 0.7\times 70\times 9.8}$

x = 1.38 m

Hence, the runner will slide to 1.38 m.

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