Question

Help! Will give medal and fan!

Answer 1

A 1,500 kg car coasts down a hill starting from rest and decreases its height by 22 m over a distance of 100 m of travel. How fast will it be going at the end of the hill, if friction has no effect in slowing its motion?

21 m/s

15 m/s

44 m/s

800 m/s

Answer 2

@perl @TheEdwardsFamily @rizzu

Answer 3

21 m/s

Answer 4

Thank you!

Answer 5

Energy is conserved or remains constant throughout.
Here the energy comes in two forms, potential energy PE and kinetic energy KE.
PE + KE = constant.

PE = mass*g * height
KE = 1/2 * mass*velocity ^2


At the top of the hill you are at rest, so the total energy is
PE + KE = mgh +1/2m*0^2 = mgh + 0 = mgh

At the bottom of the hill you have only kinetic energy .
PE + KE= m*g*0 + KE = 0 + 1/2 * mv^2 = 1/2mv^2

Since PE + KE = constant, we have

1/2 mv^2 = mgh

solve this algebraically
cancel mass m

v^2 = 2gh
v = sqrt( 2gh)

Now we can plug in given values.
v = sqrt( 2 * 9.8 m/s^2 * 22) = 20.765 m/s

Answer 6

Thank you so much! You helped a lot!

Answer 7

Perl actually explained how to do it. It makes sense for the medal to go to them.