Question

A 0.351 kg particle slides around a horizontal track. The track has a smooth, vertical outer wall forming a circle with a radius of 1.53 m. The particle is given an initial speed of 8.41 m/s. After one revolution, its speed has dropped to 5.52 m/s because of friction with the rough floor of the track. Calculate the energy loss due to friction in one revolution.

Answer 1

`- 7.07` J

Step-by-step explanation:

`m` = mass of the particle = 0.351 kg

`r` = radius of circle = 1.53 m

`v_(o)` = initial speed = 8.41 m/s

`v_(f)` = final speed after one revolution = 5.52 m/s

`\Delta K` = Loss of energy due to friction

Here we use Work-change in kinetic energy theorem to calculate the change in kinetic energy which is also the energy lost by the particle

`\Delta K = (0.5) m (v_(f)^(2) - v_(o)^(2))`

Inserting the values

`\Delta K = (0.5) (0.351) (5.52^(2) - 8.41^(2))`

`\Delta K = - 7.07` J