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

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Answer 1 · 2018-03-11T19:14:41

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

the centripetal force on the circular track is given by

$f = m . \frac{v^{2}}{r}$ $f = m. v^2/r$ where m is mass ,r the radius and v the speed

initially it was $f 1 = 4 k g . \frac{v^{2}}{6 m}$ $f1 = 4kg. v^2 /(6m$

therefore $v^{2} = (\frac{6}{4}) f 1$ $v^2 = (6/4) f1$

The initial Kinetic energy $= (\frac{1}{2}) m . v^{2}$ $= (1/2)m. v^2$ = 16 joule(J)

Therefore $(\frac{1}{2}) .4 k g . (\frac{6}{4}) f 1 = 16 J$ $(1/2) .4kg. (6/4) f1 = 16J$

so f1 the initial centripetal force $= \frac{16}{3} N$ $= 16/3 N$

now if the kinetic energy is increased to 80 J the train will

speed up and $f 2 = (\frac{4}{6}) . v e l .^{2}$ $f2 = (4/6 ).vel. ^2$

thereby $(\frac{1}{2}) m . v e l .^{2} = 80 J$ $(1/2)m.vel.^2 = 80 J$

putting up value of velocity ,we get

                  #(1/2). 4kg.( 6/4 ). f2 = 80J#

                                        #f2= 80/3 N#

so the change in centripetal force is 4 times the initial value.

A model train, with a mass of 4kg4 kg, is moving on a circular track with a radius of 6m6 m. If the train's kinetic energy changes from 16j16 j to 80j80 j, by how much will the centripetal force applied by the tracks change by?