Question

A stone of mass 2kg is falling from rest from top of a steep hill. What will be it's K.E. after 5 sec ? (g=10m/s)

Answer 1

Here,

u=0 (because initially the object was in rest)

Lets take a or g=9.8m/s^2

Now, v=u+at (first equation of motion)

= v= 0+ 9.8*5

=v =49m/s

Now, kinetic energy=1/2mv^2

1/2*2*49*49

=2401 joule

Therefore, kinetic energy =2401 joule

Answer 2

Given,

Mass of a stone = 2 Kg

The stone is falling from rest, from the top of a steep hill.

To find,

The kinetic energy of the stone after 5 seconds of its drop.

Solution,

We can simply solve this numerical problem by using the following process:

Let us assume that the velocity of the falling stone after 5 seconds is v m/s.

Mathematically,

Mathematically,I) final velocity = (initial velocity) + acceleration×time taken

Mathematically,I) final velocity = (initial velocity) + acceleration×time takenII) Kinetic energy of a body = 1/2 × (mass of the body) × (velocity of the body during the count of the energy)^2

Mathematically,I) final velocity = (initial velocity) + acceleration×time takenII) Kinetic energy of a body = 1/2 × (mass of the body) × (velocity of the body during the count of the energy)^2{Equation-1}

Now, according to the question;

the initial velocity of the falling stone = 0 m/s

acceleration of the stone = acceleration due to gravity of the earth = 10 m/s^2

So, the final velocity of the falling stone after 5 seconds of drop

= (initial velocity) + acceleration×time taken

(initial velocity) + acceleration×time taken{according to the equation-1 (I)}

= 0 m/s + (10 m/s^2 × 5s)

= 50 m/s

Now, the kinetic energy of the stone after 5 seconds of its drop

= 1/2 × (mass of the stone) × (velocity of the stone after 5 seconds of its drop)^2

{according to the equation-1 (II)}

= 1/2 × 2 Kg × (50 m/s)^2

= 2500 Kg.m^2/s^2 = 2500 J

Hence, the kinetic energy of the stone after 5 seconds of its drop is equal to 2500 J.