Question

A 0.358 kg particle slides around a horizontal track. The track has a smooth, vertical outer wall forming a circle with a radius of 1.66 m. The particle is given an initial speed of 8.19 m/s. After one revolution, its speed has dropped to 6.29 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:

The energy loss due to friction in one revolution is 4.92 J.

Explanation:

Given that,

Mass of particle = 0.358 kg

Radius = 1.66 m

Initial speed =8.19 m/s

Final speed = 6.29 m/s

We need to calculate the kinetic energy

Using formula of kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{f}^2-v_{i}^2)$

Put the value into the formula

$\Delta K.E=\dfrac{1}{2}\times0.358\times(8.19^2-6.29^2)$

$\Delta K.E=4.92\ J$

Hence, The energy loss due to friction in one revolution is 4.92 J.