A rollercoaster stops at a point with GPE of 10kJ and then travels down a frictionless slope reaching a speed of 10 m/s at ground level. After this, what length of horizontal track (friction coefficient = 0.5) is needed to bring the rollercoaster to rest?

  • Google+ icon
  • LinkedIn icon
  • 596 views

Recognise that the intial potential energy and kinetic energy at 10 m/s position should be identical due to the frictionless slope. 

mgh = 0.5mv2

10kJ = 0.5 x m x 102

10 000 = 0.5 x m x 100

50m = 10 000

m = 200 kg

Recognise that the weight of the rollercoaster is 200g which is equivalent to the vertical reaction force on the horizontal track.

The frictional force is given by the coefficient of friction multiplied by the vertical reaction force:

F = 200g 0.5 = 100g

The rollercoaster comes to rest when its energy is zero and all of the initial kinetic energy (at 10 m/s) has been dissipated by the frictional force. Therefore, we can write the work done by friction, W, in terms of the length of horizontal track, L and equate this to the kinetic energy:

W = L = 100g L = 10 000

L = 10 000 / 100g = 100 / g = 10.2m

The length of track needed is 10.2m

Daniel M. A Level Maths tutor, GCSE Maths tutor

About the author

is an online A Level Maths tutor with MyTutor studying at Durham University

Still stuck? Get one-to-one help from a personally interviewed subject specialist.

95% of our customers rate us

Browse tutors

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok