Question

A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike rider is 91kg. What is the kinetic energy of the bike and the rider just before riding up the hill? if the rider just coasts up the hill, at what height will the bike come to a stop?(neglect friction)

Answer 1

a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):

$KE = \frac{1}{2}m* v^{2}$

It is given:

v = 8.5 m/s

m = 91 kg

So:

$KE= \frac{1}{2}*91*8.5^{2} =3,287.4J$

b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h

In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J

m = 91 kg

g = 9.81 m/s² - gravitational acceleration

h = ? - height

Now, let's replace those:

3,287.4= 91 * 9.81 * h

⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m