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A bartender slides a mug of beer along a bar to a customer. The mug starts with a speed of 1.00 m/s, but by the time it reaches the customer’s hand, it has slowed to 0.25 m/s. If the filled mug has a mass of 0.850 kg, how much work is done on the mug by kinetic friction as it slides along the bar to the customer?

Respuesta :

Answer:

Explanation:

Initial velocity u = 1 m /s

final velocity v = .25 m /s

mass of the mug m = .85

initial kinetic energy

= 1/2 x .85 x 1 x 1

= .425 J

final kinetic energy

= 1/2 x .85 x .25 x .25

= .0265 J

Loss of kinetic energy

= .425 - .0265 J

= .3985 J

Work done by friction

= - .3985 J  ( loss of kinetic energy is due to negative work by friction )

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