Question

I NEED CORRECT ANSWER.
A body of mass 10 kg is kept at a height of 5 m. It is allowed to fall and reach the ground.
(i)What is the total mechanical energy possessed by the body at the height of 2 m assuming it is a frictionless medium.
(i)What is the kinetic energy possessed by the body just before hitting the ground? Take g=10m/s2.

Answer 1

A body of mass 10 kg is kept at a height of 5 m. It is allowed to fall and reach the ground.  

(i)What is the total mechanical energy possessed by the body at the height of 2 m assuming it is a frictionless medium.  

(i)What is the kinetic energy possessed by the body just before hitting the ground? Take g=10m/s2.

Mass = 10 kg

Height = 5 m

g = 10 m/s²

Mechanical Energy = Potential Energy + Kinetic Energy

Potential Energy at h of 5 m =  mgh =  10 * 10 * 5 = 500

Kinetic energy at h of 5m  = 0  as body is at rest (KE = (1/2)mV² and V=0) so KE = 0)

Total Mechanical Energy = 500 + 0 = 500 J

i)

Energy neither can be created or destroyed but can be converter in to another form. as its a Frictionless surface so no energy loss in any other form

so

Total Mechanical Energy at h of 2m = 500  J

ii)

Just Before hitting the ground  h = 0 m

so potential energy = 0 J

Total Mechanical Energy = 500 J

500 = 0 + Kinetic Energy

=> Kinetic Energy = 500 J

or we can find it by using

KE = (1/2)mV²

V² - U² = 2aS   U = 0  S = 5 ,  a = g = 10m/s²

V² = 2*10*5= 100

KE = (1/2)*10 * 100 = 500 J

in part i) if we need to found  Kinetic energy & Potential energy separately

Potential energy at 2 m = 10 * 10 * 2 = 200 J

Kinetic Energy = Total energy - potential energy = 500 - 200 = 300 J

or we can calculate by

Kinetic Energy = (1/2) m V²

2aS = V² - U²

U = 0  , S = 5 -2 = 3 m  a = g = 10m/s²

=> 2*10*3 = V²

=> V² = 60

Kinetic Energy = (1/2) 10 * 60 = 300 J

Answer 2

Answer:

(i): Total mechanical energy of the body at height 2 m = 490 J.

(ii): Kinetic energy of the body just before hitting the ground = 490 J.

Explanation:

Given:

• Mass of the body, $m = 10\ kg$.

• Height from where the body is allowed to fall, $h = 5\ m$.

Assumptions:

• $u$ = initial velocity of the body at height 5 m.

• $v_1$ = velocity of the body at height 2 m.

• $v$ = velocity of the body just before hitting the ground.

• $a$ = acceleration of the body.

• $g$ = acceleration due to gravity = $9.8\ m/s^2$.

Since, the body falls from a height, therefore the initial velocity of the body at this height is 0 i.e., $u=0\ m/s$.

The body is falling freely it means that the acceleration acting on the body is equal to the acceleration due to gravity.

$a=g=9.8\ m/s^2$.

Using the Kinematics equation,

$v^2-u^2=2aS\ .......(1).$

where,

$S$ = distance traveled by the body.

Part (i):

For the height 2 m,

The distance traveled by the body, from the height 5 m, is $S=5-2=3\ m$ and $v=v_1$

Putting the value for height 2 m in the above equation (1),

$v_1^2-u^2=2aS\\v_1^2-0^2=2\times 9.8\times 3\\v_1^2 = 58.8\\v_1 = 7.668\ m/s.$

The kinetic energy of the body at height 2 m is given by

$K_1 = \dfrac 12 mv_1^2\\=\dfrac 12 \times 10\times (7.668)^2\\=294\ J.$

The potential energy of the body at this height is given by

$U_1 = mgh_1=10\times 9.8\times 2=196\ J.$

$h_1$ is height 2m.

Thus, the total mechanical energy at this height is given by

$E_1 = K_1+P_1\\=294+196\\=490\ J.$

Part (ii):

For the ground level,

Using equation (1), $S = 5-0=5\ m$.

$v^2-u^2=2g\cdot 5 \\v^2=2\times 9.8\times 5 = 98\\v = 9.899\ m/s.$

So, the kinetic energy is given by

$K_2 = \dfrac 12 mv ^2 \\=\dfrac 12 \times 10\times (9.899)^2\\=490\ J.$