# A model train with a mass of 4 kg is moving along a track at 21 (cm)/s. If the curvature of the track changes from a radius of 84 cm to 140 cm, by how much must the centripetal force applied by the tracks change?

Aug 29, 2017

$8400 \mathrm{dy} n e \mathmr{and} 0.084 N$

#### Explanation:

$m = 4000 g$

$v = 21 c m \cdot {s}^{- 1}$

${R}_{1} = 84 c m$

${F}_{c \left(1\right)} = \frac{m {v}^{2}}{R} _ 1$

$= \frac{4000 \cdot {21}^{2}}{84}$

$= 21000 \text{ dyne}$

${R}_{2} = 140 c m$

${F}_{c \left(2\right)} = \frac{m {v}^{2}}{R} _ 2$

$= \frac{4000 \cdot {21}^{2}}{140}$

$= 12600 \text{ dyne}$

Change in ${F}_{c} = {F}_{c \left(1\right)} - {F}_{c \left(2\right)}$

=21000-12600 " dyne" =8400" dyne" =0.084 N