Question

A roller coaster starts at rest on the slope of a track , 26 m above the ground . it coasts down that slope and then climbs another , at which point it is 16 m above the ground . how fast is it moving at the top of this second slope? ignore friction .

Answer 1

Given:

A roller coaster starts at rest on the slope of a track , 26 m above the ground . it coasts down that slope and then climbs another , at which point it is 16 m above the ground .

To find:

Speed at the top of 2nd slope ?

Calculation:

In this kind of questions, it is best to apply CONSERVATION OF MECHANICAL ENERGY PRINCIPLE :

$\sf \: PE1 + KE1 = PE2 + KE2$

$\sf \implies \: mg(26) + \dfrac{1}{2} m {u}^{2} = mg(16) + \dfrac{1}{2} m {v}^{2}$

$\sf \implies \: mg(26) + \dfrac{1}{2} m {(0)}^{2} = mg(16) + \dfrac{1}{2} m {v}^{2}$

$\sf \implies \: mg(26) = mg(16) + \dfrac{1}{2} m {v}^{2}$

$\sf \implies \: \dfrac{1}{2} m {v}^{2} = mg(26 - 16)$

$\sf \implies \: \dfrac{1}{2} m {v}^{2} =10 mg$

$\sf \implies \: \dfrac{1}{2} {v}^{2} =10 g$

$\sf \implies \: {v}^{2} =20 g$

$\sf \implies \: {v}^{2} =200$

$\sf \implies \: v = \sqrt{200}$

$\sf \implies \: v = 10\sqrt{2} \: m {s}^{ - 1}$

So, Velocity at top of 2nd slope is 10√2 m/s.