Question

17. A body is released from position A as shown in
figure. The speed of body at position B is
[NCERT Pg. 121]
А.
50 m
-В
10 m
(1) 10 m/s
(2) 10/2 m/s
(3) 20 m/s
(4) 2012 m/s​

Answer 1

CORRECT QUESTION

Body is released from position A as shown in

figure. The speed of body at position B is

(1) 10 m/s

(2) 10/2 m/s

(3) 20 m/s

(4) 20√2 m/s

ANSWER

The answer is 20√2 m/s

GIVEN

Body is released from position A

TO FIND

The speed of body at B

SOLUTION

We can simply solve the above problem as follows;

We know that,

Potential energy at point A = mgh₁

Where,

m = mass of the ball

g = acceleration due to gravity

h₁ = 50 meters

Therefore,

Energy at A = 50mg

Potential energy at B = mgh₂ = 10mg

Also,

The ball has kinetic energy at point B

Kinetic energy at point B = 1/2 × mv²

Where,

v = velocity of the ball.

Also,

Energy at point A = energy at point B

Therefore,

mgh₁ = mgh₂ + 1/2 × mv²

50mg = 10mg + 1/2 × mv²

1/2× mv² = 40mg

v² = 40 × 10 × 2

v² = 800

v = √800 = 20√2 m/s

Hence, The answer is 20√2 m/s

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Answer 2

The speed of the body at position B is 20√2 m/s

As per given The body is released from position A

Now here we have to find The speed of the body at B

Therefore

The above problem is easily resolved as shown below;

We know that,

The potential energy at point A = mgh₁

Where,

m = mass of the ball

g = acceleration due to gravity

h₁ = 50 meters

Therefore,

Energy at A = 50mg

Potential energy at B = mgh₂ = 10mg

Also,

At point B, the ball has kinetic energy.

Kinetic energy at point B = $\frac{1}{2}$ × mv²

Where

v = velocity of the ball.

Also,

The energy at point A = energy at point B

Therefore,

$mgh_{1} = mgh_{2} + \frac{1}{2} *mv^{2}$

$50mg = 10mg + \frac{1}{2} *mv^{2}$

$\frac{1}{2} * mv^{2} = 40mg$

$V^{2}$  = 40 × 10 × 2

$V^{2}$  = 800

V   = √800

V   = 20√2 m/s

The answer is 20√2 m/s

Hence, The speed of the body at position B is 20√2 m/s

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