During an emergency stop, a #1.5*10^3#-kilogram car lost a total of #3.0 * 10^5# joules of kinetic energy. What was the speed of the car at the moment the brakes were applied?

1 Answer
Steve J.
Jun 11, 2018

#v = 20 m/s#

Explanation:

The car had #3.0*10^5 J# of kinetic energy before the brakes were applied. The formula for kinetic energy is

#KE = 1/2*m*v^2#

plugging in the data,
#3.0*10^5 J = 1/2*1.5*10^3 kg*v^2#

#v^2 = (6.0*10^5 J)/(1.5*10^3 kg) = 4*10^2 J/(kg)#

How do you deal with the units of J/kg when the answer should be a velocity? The formula for KE is
#KE = 1/2*m*v^2#. KE works out to be in Joules only if m is in kg and v is in m/s. Therefore,

#v = sqrt(400) m/s = 20 m/s#

I hope this helps,
Steve

Related questions
Impact of this question
10599 views around the world
You can reuse this answer
Creative Commons License