A model train, with a mass of #5 kg#, is moving on a circular track with a radius of #9 m#. If the train's kinetic energy changes from #70 j# to #40 j#, by how much will the centripetal force applied by the tracks change by?

The centripetal force is

The kinetic energy is

The variation of kinetic energy is

The variation of centripetal force is

To calculate the change in the centripetal force applied by the tracks, we first need to find the initial and final velocities of the model train using the given kinetic energies.

The formula for kinetic energy (KE) is:

$KE = \frac{1}{2} m v^2$

Given:

• Mass (m) = 5 kg
• Initial KE = 70 J
• Final KE = 40 J
• Radius (r) = 9 m

First, calculate the initial velocity (v_initial) using the initial kinetic energy:

$70 = \frac{1}{2} \times 5 \times v_{\text{initial}}^2$ $140 = 5 \times v_{\text{initial}}^2$ $v_{\text{initial}}^2 = \frac{140}{5}$ $v_{\text{initial}}^2 = 28$ $v_{\text{initial}} = \sqrt{28}$ $v_{\text{initial}} = \sqrt{4 \times 7}$ $v_{\text{initial}} = 2\sqrt{7}$

Now, calculate the final velocity (v_final) using the final kinetic energy:

$40 = \frac{1}{2} \times 5 \times v_{\text{final}}^2$ $80 = 5 \times v_{\text{final}}^2$ $v_{\text{final}}^2 = \frac{80}{5}$ $v_{\text{final}}^2 = 16$ $v_{\text{final}} = \sqrt{16}$ $v_{\text{final}} = 4$

The centripetal force (F_c) can be calculated using the formula:

$F_c = \frac{m \times v^2}{r}$

So, the initial centripetal force (F_c_initial) is:

$F_{c_{\text{initial}}} = \frac{5 \times (2\sqrt{7})^2}{9}$ $F_{c_{\text{initial}}} = \frac{5 \times 28}{9}$ $F_{c_{\text{initial}}} = \frac{140}{9}$

And the final centripetal force (F_c_final) is:

$F_{c_{\text{final}}} = \frac{5 \times 4^2}{9}$ $F_{c_{\text{final}}} = \frac{5 \times 16}{9}$ $F_{c_{\text{final}}} = \frac{80}{9}$

The change in centripetal force is:

$\Delta F_c = F_{c_{\text{final}}} - F_{c_{\text{initial}}}$ $\Delta F_c = \frac{80}{9} - \frac{140}{9}$ $\Delta F_c = \frac{-60}{9}$ $\Delta F_c = -\frac{20}{3}$

Therefore, the centripetal force applied by the tracks will decrease by $\frac{20}{3}$ N.