What is the kinetic energy of an object with a mass of #7 kg# that has been in freefall for #12 s#?

2 Answers
Feb 13, 2016

Answer:

#4.85xx10^4 J#

Explanation:

Let's start by writing down what we know, and where we want to get to in the hopes that doing so will help us to understand what we are missing:

we know that

#m = 7kg#
#t = 12s#

and we want to know the kinetic energy which is given by

#K.E. = 1/2 m v^2#

We know the mass, now we need to know the velocity. We know that the object is in freefall, so it is experiencing acceleration due to gravity, which we know (assuming that the free fall it is experiencing is here on Earth):

#g=9.81m//s^2#

The velocity of an object under a constant acceleration increases with time as given by the following equation where we have substituted #g# for the acceleration:

#v(t) = v(t=0) + g*t#

Assuming that it started with zero velocity (the question implies this by telling us that only that the object experiences free fall)

#v(t) = 0 + 9.81m//s^2 * 12s = 118m//s#

Substituting this into the kinetic energy formula above we get:

#K.E. = 1/2 * 7kg * (118m//s)^2 = 48500 kg* m^2 //s^2#
# = 4.85xx10^4 J#

Feb 13, 2016

Answer:

There are (at least) two ways to approach this question. Both are briefly outlined below. The kinetic energy, #E_k=48.4# #kJ#.

Explanation:

First approach:

We can calculate the velocity of the object and then use that and its mass to find its kinetic energy.

The initial velocity, #u# can be assumed to be #0# #ms^-1# and the acceleration due to gravity is #9.8# #ms^-2# so we can use:

#v=u+at=0+9.8*12=117.6# #ms^-1#

Then the kinetic energy is:

#E_k=1/2mv^2=1/2*7*117.6^2=48404# #J = 48.4# #kJ#

Second approach:

All of the kinetic energy of the object has come from the change in its gravitational potential energy. The potential energy difference will be #Delta E_p=mgDeltah#. (#Delta# just means 'the change in'.) We need to find the distance it has traveled:

#Deltah=d=ut+1/2at^2=0*12+1/2*9.8*12^2=705.6# #m#

Now:

#E_k=DeltaE_p=mgDeltah=7*9.8*705.6=48404# #J# = #48.4# #kJ#

Always reassuring when both methods yield the same answer. ;-)