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dybincka [34]
2 years ago
11

A 66-kg base runner begins his slide into second base when he is moving at a speed of 3.4 m/s. The coefficient of friction betwe

en his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base(a) How much mechanical energy is lost due to friction acting on the runner. (b) How far does he slide?
Physics
1 answer:
Helen [10]2 years ago
5 0

Answer:

a ) = 381.48 J

b )=  84.25 cm

Explanation:

Kinetic energy of the runner

= 1/2 m v²

= .5 x 66 x 3.4²

= 381.48 J

The final kinetic energy of the runner is zero .

Loss of mechanical energy

= 381.48 J

This  loss in  mechanical energy is due to action of frictional force .

b )

Let s be the distance of slide

deceleration due to frictional force

= μmg/m

.7 x 66 x 9.8 / 66

a = - 6.86 m s⁻¹

v² = u² - 2 a s

0 = 3.4² - 2x6.86 s

s = 3.4² / 2x6.86

= .8425 m

84.25 cm

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Take the car's starting position to be at t_0=0\,\mathrm s. Then

\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t

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You also could have first found the acceleration using the equation

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Answer:

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\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M}    \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \  \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}

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