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3 years ago

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A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

his clothes and Earth is 0.635. He slides so that his speed is zero just as he reaches the base. The acceleration of gravity is 9.8 m/s 2 . What is the magnitude of the mechanical energy lost due to friction acting on the runner? Answer in units of J.

1 answer:

SashulF [63]3 years ago

6 0

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

$\Delta W = \Delta KE$

$\Delta W = \frac{1}{2} mv^2$

Here,

m = mass

v = Velocity

Our values are given as,

$m = 79.7kg$

$v = 4.77m/s$

Replacing,

$\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2$

$\Delta W = 907J$

Therefore the mechanical energy lost due to friction acting on the runner is 907J

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