Science, asked by JAVAID40, 10 months ago

Aisha dropped an object of mass 4 Kg from a height of 10m.
The object has a speed of 0 m/s at position A and is at a height
of 10m above the ground. At position B, the object is 6m above
the ground. At position C, the object is 4m above the ground.
At point D, the object is mere picometer ( almost at 0 meter height)
above the ground. Assume negligible air resistance throughout
the motion. Use this information to fill in the table. (g=10m/s2)​

Answers

Answered by krishna1283
4

Explanation:

Gravitational potential energy and work done

If an object is lifted, work is done against the force of gravity.

When work is done energy is transferred to the object and it gains gravitational potential energy.

If the object falls from that height, the same amount of work would have to be done by the force of gravity to bring it back to the Earth’s surface.

If an object at a certain height has 2000 J of gravitational potential energy, we can say that:

2000 J of work has been done in getting the object to that height from the ground

and

2000 J of work would have to be done to bring it back to the ground.

Answered by BrainlyPrince727
0

Given :-

The object has a speed of 0 m/s at position A and is at a height

A  of 10m above the ground.

To find :-

Velocity & time taken to reach at B , C & D

Solution :-

At  t = 0    position  A  - 10 m above ground    velocity =  0

PE = mgh = 4 * 10 * 10 = 400 Joule  

KE =(1/2)mv² = 0  

At position B  - 6 m above ground distance covered  (10 - 6) = 4 m  

S = ut + (1/2)at²

u = 0  

a = g = 10  

=> 4 = (1/2)(10)t²  

=> t = √0.8  =  0.9 sec  

V = U  + at   =  9 m/s  

PE = 4 * 10 * 6 ≈ 240 joule

KE = 400 - 240 ≈ 160 Joule  

At position C  - 4 m above ground distance covered  (10 - 4) =6 m  

S = ut + (1/2)at²

u = 0  

a = g = 10  

=> 6 = (1/2)(10)t²  

=> t = √1.2  ≈ 1.1 sec  

V = U  + at   ≈  11 m/s    

PE = 4 * 10 * 4 = 160 joule  

KE = 400 - 160 = 240 Joule  

At position D  - 0 m above ground distance covered  (10 - 0) =10 m  

S = ut + (1/2)at²    

u = 0  

a = g = 10  

=> 10 = (1/2)(10)t²  

=> t = √2  = 1.4sec  

V = U  + at   =  14 m/s  

PE = 4 * 10 * 0 =  0 joule  

KE = 400  Joule  

Position  Distance from ground  PE    KE     Time         Velocity        

A                   10 m                        400   0        0  s             0 m/s  

B                   6 m                          240  160     0.9 s           9 m/s  

C                   4 m                          160   240     1.1 s            11 m/s  

D                   0 m                          0       400    1.4 s           14 m/s

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