"The Potential Energy Of A 4 Kg Object On Top Of A Hill Is 72 J. What Is Its Velocity In M/S Just Before It Hits The Ground?, A. 36, B. 18, C. 6, D. 3

July 22, 2022

"The potential energy of a 4 kg object on top of a hill is 72 J. What is its velocity in m/s just before it hits the ground?

a. 36
b. 18
c. 6
d. 3"

Answer:

The velocity in the problem wherein a 4kg object on top of the hill has a potential energy of 72J is c. 6m/s.

Explanation:

The first step that we will do is to understand the concepts of kinetic energy and potential energy. Kinetic energy is the work that an object should have in order for it to be in motion or accelerate. Simply, kinetic energy is the object's energy caused by its movement in space. The energy is invested through its acceleration and maintains its magnitude unless a change of velocity occurs. The kinetic energy of an object will be set to 0 once it is placed into rest.

On the other hand, potential energy is the energy that an object contains within itself which is dependent on its mass, distance from a relative mass or point of reference, and gravitational pull. Given its definition, your own potential energy may vary on the point of reference you are using. Example, if you are on rooftop of a 5-storey building, your potential energy will be different from each floor. Your potential energy from the rooftop in reference to the 5th floor will be different from your potential energy if your reference is the 2nd floor.

The formula for the kinetic energy is $KE = \frac{1}{2}mv^2$

The formula for the potential energy is $PE = mgh$

Next, we will answer the problem in GRESA (Given, Required, Equation, Solution, Answer) format for an easier view on the problem and its solutions.

The potential energy of a 4kg object on top of a hill is 72J. What is its velocity in m/s just before it hits ground.

GIVEN

m = 4kg

PE = 72J

REQUIRED

v = ?

EQUATION

$PE = mgh\\{v_f}^2=v_o+2ad$

SOLUTION

In order to solve the problem, we need to investigate the formula for velocity. The initial velocity will be 0 since we started from rest. The g stands for gravity which is . We notice that we only lack the magnitude of distance so we will start from there using the potential energy formula.

$PE = mgh\\\\72J = (4kg)(9.8 m/s^2)h\\\\h=\frac{72J}{(4kg)(9.8 m/s^2)}\\\\h=1.84m\\\\{v_f}^2=v_o+2ad\\\\{v_f}^2=0+2(9.8m/s^2)(1.84m)\\\\{v_f}=\sqrt{36.064m^2s^2}\\\\{v_f}= 6 m/s$