A model train, with a mass of #4 kg#, is moving on a circular track with a radius of #6 m#. If the train's kinetic energy changes from #16 j# to #80 j#, by how much will the centripetal force applied by the tracks change by?

Question

A model train, with a mass of #4 kg#, is moving on a circular track with a radius of #6 m#.  If the train's kinetic energy changes from #16 j# to #80 j#, by how much will the centripetal force applied by the tracks change by?

Leo Abeyta · Accepted Answer

the centripetal force changes by four times of the initial value.

the centripetal force on the circular track is given by #f = m. v^2/r# where m is mass ,r the radius and v the speed initially it was # f1 = 4kg. v^2 /(6m # therefore #v^2 = (6/4) f1# The initial Kinetic energy # = (1/2)m. v^2 # = 16 joule(J) Therefore # (1/2) .4kg. (6/4) f1 = 16J # so f1 the initial centripetal force # = 16/3 N # now if the kinetic energy is increased to 80 J the train will speed up and # f2 = (4/6 ).vel. ^2# thereby # (1/2)m.vel.^2 = 80 J # putting up value of velocity ,we get so the change in centripetal force is 4 times the initial value.

Leo Abeyta · Answer

undefined To find the change in centripetal force applied by the tracks, you can use the formula relating kinetic energy to centripetal force:

$$KE = \frac{1}{2}mv^2 = \frac{1}{2}m(\frac{r^2}{r})^2$$

where:
- $KE$ is the kinetic energy,
- $m$ is the mass of the object,
- $v$ is the velocity,
- $r$ is the radius of the circular path.

First, find the initial velocity ($v_1$) using the initial kinetic energy ($KE_1$) and mass ($m$):

$$KE_1 = \frac{1}{2}m{v_1}^2$$

Then, find the final velocity ($v_2$) using the final kinetic energy ($KE_2$) and mass ($m$):

$$KE_2 = \frac{1}{2}m{v_2}^2$$

Once you have $v_1$ and $v_2$, use the formula for centripetal force:

$$F = \frac{mv^2}{r}$$

Subtract the initial centripetal force ($F_1$) from the final centripetal force ($F_2$) to find the change:

$$\Delta F = F_2 - F_1$$

Substituting the given values, solve for the change in centripetal force.