Question

A 4768kg roller coaster train full of riders approaches the landing dock at a speed of 17.1m/s. It is abruptly decelerated to a speed of 2.2m/s over a distance of 13.6m. What is the magnitude of the retarding force which acts upon the roller coaster cars?

Answer 1

`F ~~50,409` N

Explanation:

Given: roller coaster with `m = 4768` has a speed `= 17.1`m/s. It abruptly decelerated to a speed `= 2.2` m/s over a distance of `13.6` m.

To find the magnitude of the retarding force we need to find the initial kinetic energy and the final kinetic energy. The external work done to retard the train acts on the initial energy of the train.

This is the equation for this situation:

`KE_i + W_(ext) = KE_f`

Use these formulas: `" "KE = 1/2 mv^2;`

`W_(ext) = Fd cos 180^@ = F (13.6) (-1) = -13.6F`

Solve for the Force: `" "F = (KE_f - KE_i)/-13.6 = (m(v_f^2 - v_i^2))/(-2*13.6)`

`KE_i = 1/2 * 4768 kg * (17.1 m/s)^2 = 697,105.44 " Joules"`

`KE_f = 1/2 * 4768 kg * (2.2 m/s)^2 = 11,538.56 " Joules"`

`F = (4768(2.2^2 - 17.1^2))/(-2*13.6) ~~ 50,409.33 " N"`