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

Answer:

$- 7.07$ J

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