Question

aisha dropped an object of mass 4kg from. a height of 10m.the object has a speed of 0m/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 metre height) above the ground. assume negligible air resistance through out the motion. use the information to fill in the table (g=10m/s²) ​

Answer 1

According to the question,

Object's speed at position A = 0 m/sec

Height from where object is dropped to the ground = 10 m

Now,

(A) When t = 0, position  A  = 10 m, and velocity =  0 m/sec  

Then, PE = m×g×h J = (4 × 10 × 10) J = 400 J

And, KE = $\frac{1}{2}mv^2 = 0$

(B) position B = 6 m, and distance covered = (10 - 6) m = 4 m

Then, $S = ut + \frac{1}{2}at^2$

We have,

u = 0

a = g = 10

So,

4 = $(\frac{1}{2})(10)t^2$

=> t = $\sqrt{0.8}$ sec = 0.9 sec

Then, $V = U + at =$ 9 m/sec

So, PE = (4 × 10 × 6) J = 240 J

KE = (400 - 240) J = 160 J

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

In the same manner ,  $S = ut + \frac{1}{2}at^2$

We have,

u = 0

a = g = 10

so, 6 =  $(\frac{1}{2})(10)t^2$

=> t = $\sqrt{1.2}$ secs = 1.1 sec s

Then, $V = U + at =$ 11 m/sec

Then, PE = (4 × 10 × 4) J = 160 J

And, KE = (400 - 160) J = 240 J

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

 In the same manner ,  $S = ut + \frac{1}{2}at^2$

We have,

u = 0

a = g = 10

so, 10 =  $(\frac{1}{2})(10)t^2$

=> t = $\sqrt{2}$ = 1.4 secs

Then, $V = U + at =$ 14 m/sec

Then, PE = (4 × 10 × 0) J = 0 J

And, KE = (400 - 0) J = 400 J

The progression of all the derived values have been organised in a tabular manner given below.

A similar answer is available at - https://brainly.in/question/21663307

Answer 2

Given :   aisha dropped an object of mass 4kg from. a height of 10m.the object has a speed of 0m/s at position A and is at a height of 10m above the ground

To Find :  PE  , KE   at A , B , C & D  

Solution:

At point A   - 10 m height v = 0

PE = mgh = 4 * 10 x 10  = 400 J

KE  = (1/2)mv² = (1/2)x 4* 0 = 0 J

TE = 400 + 0  = 400 J

At point B  - 6m

PE = mgh = 4 * 10 x 6  =240 J

KE  = TE - PE = 400 - 240 = 160J

At point C  - 4m

PE = mgh = 4 * 10 x 4  =160 J

KE  = TE - PE = 400 -160 = 240J

At point D  - 0m

PE = mgh = 4 * 10 x 0  =00 J

KE  = TE - PE = 400 -00 =  400J

Position  Distance from ground  PE (J)    KE(J)    

A                   10 m                       400       0      

B                   6 m                         240     160    

C                  4 m                          160      240    

D                   0 m                        0          400      

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