What is the kinetic energy of an object with a mass of #9 kg# that has been in freefall for #1 s#?

2 Answers
May 26, 2016

Answer:

The kinetic energy is approximately 433 J.

Explanation:

The kinetic energy, in classical mechanics, it is

#E=1/2mv^2#, where #m# is the mass and #v# is the velocity.
You have the mass, so you need to calculate the velocity.
Your body is in free fall, this means that it is subject to the gravity that accelerates with a constant strength.

The acceleration is the change of velocity during the time. So we can write: #a=(v_f-v_i)/(t_f-t_i)#. In our case the initial conditions are that both velocity and time are zero when the body starts to fall.
Then #v_i=0, t_i=0#. We have #a=v_f/t_f# and we are interested in the #v_f# that is #v_f=a*t_f#. The #t_f=1# s, the acceleration is not given but you probably know from some of your lectures that the acceleration of gravity on Earth is approximately #9.81# #m/s^2#.

#v_f=9.81# #m/s^2# #1# #s# #= 9.81# #m/s#.
With the velocity we can calculate the energy

#E=1/2 9# kg #(9.81 m/s)^2=1/2 9*96.2361 \approx 433# J.

May 26, 2016

Answer:

I found: #432.2J=K#

Explanation:

We can think of our system in which the object was placed at a certain height and then released. Now, when it was placed at the initial position it gained a certain Potential Energy #U# that was entirely converted into Kinetic Energy #K# during the fall giving: #U=K#!!!

We know that:
Potential Energy: #U=mgh#
where:
#m=# mass;
#g=# acceleration of gravity;
#h=y_f-y_i=# height.

From kinematics we have:
#h=y_f-y_i=v_it+1/2at^2#
with:
#g=9.8m/s^2# DOWNWARDS;
#v_i=0# is the initial velocity;

so:

#y_f-y_i=(0*1)-(1/2)9.8(1^2)#
considering the ground as #y_f=0#

#h=y_i=4.9m#
and finally:
#U=mgh=9*9.8*4.9=432.18~~432.2J=K#