A body falls freely through a height of 5m .find the kenetic energy when it reach the ground and slow that it equals the potential energy before falling?

Question

10 kg body falls freely through height of 5m .what is the Kenetic energy when it reach the ground and the potential energy before falling?

 

Answer 1

Arturo's solution definitely hits the mark.  But, as I read the question, it sounds like it may want you to demonstrate that the kinetic energy at the ground equals the potential energy at the initial height, without assuming conservation of mechanical energy (and assuming the object is dropped from rest; otherwise, it would also have kinetic energy at the start, which would affect its speed and kinetic energy at the ground)

 

We can calculate the potential energy the object has right from the start.  A 10 kg object at 5 m above the ground (our usual zero-potential-energy level) has a gravitational potential energy, as Arturo showed, of:

 

GPE = 490 J

 

We can determine the kinetic energy at the ground by determining the speed at the ground.  For an object free falling down through 5 m, we can set up a kinematic equation, where we have:

 

to find:  final velocity

know:  acceleration (= -9.8 m/s², taking down as negative), initial velocity (= 0, starts at rest), and displacement (= -5 m, since it falls in what I have called the negative direction)

 

So we can use the kinematic equation:

 

v² = v_o²+2a(x-x_o)

 

where

v = final velocity (what we want to solve for)

v_o = initial velocity

a = acceleration

(x-x_o) = displacement

 

So, putting in the values we know, we have:

 

v² = (0 m/s)²+2(-9.8 m/s²)(-5 m)

 

v² = 98

 

Actually, since all we really need to calculate kinetic energy is the magnitude of final velocity squared, this number will be all we need.

 

KE_ground = (1/2)mv² = (1/2)(10 kg)(98) = 490 J

 

This is the exact same value as the initial potential energy.

 

In general, using that formula, you end up with:  v² = 2gh

 

Then, KE = (1/2)m(v²) = (1/2)m(2gh) = mgh, which is just the potential energy of an object at height h. 

 

Answer 2

P = potential energy

 

P = mgh

 

K = kinetic energy

 

K = (1/2)mv²

 

Energy conservation:

 

At different points 1 and 2,

 

P₁ + K₁ = P₂ + K₂

 

mgh₁ + (1/2)mv₁² = mgh₂ + (1/2)mv₂²

 

Before falling, h₁ = 5m, v₁ = 0

At the bottom of the fall, h₂ = 0, v₂ = ?

 

mgh₁ = (1/2)mv₂² = K₂

 

K₂ = P₁ = mgh₁ = (10 kg)(9.8 m/s²)(5m) = 490 J