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Work, Energy and Power

268662 A cradle is ' $h$ ' meters above the ground at the lowest position and ' $H$ ' meters when it is at the highest point. If ' $v$ ' is the maximum speed of the swing of total mass ' $m$ ' the relation between ' $h$ ' and ' $H$ ' is

1

(v^{2} / 2 g) + h = H

2

(v^{2} / g) + 2 h = H

3

(v^{2} / 2 g) + H = h

4

1 / 2 m v^{2} + h = H

Work, Energy and Power

268663 $A B$ is a frictionless inclined surface making an angle of $30^{\circ}$ with horizontal. A is $6.3 m$ above the ground while $B$ is $3.8 m$ above the ground. $A$ block slides down from $A$ , initially starting from rest. Its velocity on reaching $B$ is
![original image](https://cdn.mathpix.com/snip/images/DkkuvOboa_of-qrZXfk7iFbnlFFTxaIfbcZBLFuvRdQ.original.fullsize.png)

1

14.14 m / s

2

7.07 m / s

3

5 m / s

4

25 m / s

Explanation:

gain in $K . E =$ Loss of P.E $= m g (h_{1} - h_{2})$

Work, Energy and Power

268664 A stone of mass " $m$ " initially at rest and dropped from a height " $h$ " strikes the surface of the earth with a velocity " $v$ ". If the gravitational force acting on the stone is $W$ , then which of the following identities is correct

1

mv - mh = 0

2

1 / 2 {mv}^{2} - {Wh}^{2} = 0

3

1 / 2 m v^{2} - W h = 0

4

1 / 2 m v^{2} - m h = 0

Explanation:

gain in $K . E =$ Loss of P.E

Work, Energy and Power

268724 A block of mass $4 kg$ slides on a horizontal frictionless surface with a speed of $2 m / s$ . It is brought to rest in compressing a spring in its path. If the force constant of the spring is 400 $N / m$ , by how much the spring will be compressed

1

2 \times 10^{- 2} m

2

0.2 m

3

20 m

4

200 m

Explanation:

K.E of block is converted into elastic P.E in spring i.e $\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

Work, Energy and Power

268662 A cradle is ' $h$ ' meters above the ground at the lowest position and ' $H$ ' meters when it is at the highest point. If ' $v$ ' is the maximum speed of the swing of total mass ' $m$ ' the relation between ' $h$ ' and ' $H$ ' is

1

(v^{2} / 2 g) + h = H

2

(v^{2} / g) + 2 h = H

3

(v^{2} / 2 g) + H = h

4

1 / 2 m v^{2} + h = H

Work, Energy and Power

268663 $A B$ is a frictionless inclined surface making an angle of $30^{\circ}$ with horizontal. A is $6.3 m$ above the ground while $B$ is $3.8 m$ above the ground. $A$ block slides down from $A$ , initially starting from rest. Its velocity on reaching $B$ is
![original image](https://cdn.mathpix.com/snip/images/DkkuvOboa_of-qrZXfk7iFbnlFFTxaIfbcZBLFuvRdQ.original.fullsize.png)

1

14.14 m / s

2

7.07 m / s

3

5 m / s

4

25 m / s

Explanation:

gain in $K . E =$ Loss of P.E $= m g (h_{1} - h_{2})$

Work, Energy and Power

268664 A stone of mass " $m$ " initially at rest and dropped from a height " $h$ " strikes the surface of the earth with a velocity " $v$ ". If the gravitational force acting on the stone is $W$ , then which of the following identities is correct

1

mv - mh = 0

2

1 / 2 {mv}^{2} - {Wh}^{2} = 0

3

1 / 2 m v^{2} - W h = 0

4

1 / 2 m v^{2} - m h = 0

Explanation:

gain in $K . E =$ Loss of P.E

Work, Energy and Power

268724 A block of mass $4 kg$ slides on a horizontal frictionless surface with a speed of $2 m / s$ . It is brought to rest in compressing a spring in its path. If the force constant of the spring is 400 $N / m$ , by how much the spring will be compressed

1

2 \times 10^{- 2} m

2

0.2 m

3

20 m

4

200 m

Explanation:

K.E of block is converted into elastic P.E in spring i.e $\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

Work, Energy and Power

268662 A cradle is ' $h$ ' meters above the ground at the lowest position and ' $H$ ' meters when it is at the highest point. If ' $v$ ' is the maximum speed of the swing of total mass ' $m$ ' the relation between ' $h$ ' and ' $H$ ' is

1

(v^{2} / 2 g) + h = H

2

(v^{2} / g) + 2 h = H

3

(v^{2} / 2 g) + H = h

4

1 / 2 m v^{2} + h = H

Work, Energy and Power

268663 $A B$ is a frictionless inclined surface making an angle of $30^{\circ}$ with horizontal. A is $6.3 m$ above the ground while $B$ is $3.8 m$ above the ground. $A$ block slides down from $A$ , initially starting from rest. Its velocity on reaching $B$ is
![original image](https://cdn.mathpix.com/snip/images/DkkuvOboa_of-qrZXfk7iFbnlFFTxaIfbcZBLFuvRdQ.original.fullsize.png)

1

14.14 m / s

2

7.07 m / s

3

5 m / s

4

25 m / s

Explanation:

gain in $K . E =$ Loss of P.E $= m g (h_{1} - h_{2})$

Work, Energy and Power

268664 A stone of mass " $m$ " initially at rest and dropped from a height " $h$ " strikes the surface of the earth with a velocity " $v$ ". If the gravitational force acting on the stone is $W$ , then which of the following identities is correct

1

mv - mh = 0

2

1 / 2 {mv}^{2} - {Wh}^{2} = 0

3

1 / 2 m v^{2} - W h = 0

4

1 / 2 m v^{2} - m h = 0

Explanation:

gain in $K . E =$ Loss of P.E

Work, Energy and Power

268724 A block of mass $4 kg$ slides on a horizontal frictionless surface with a speed of $2 m / s$ . It is brought to rest in compressing a spring in its path. If the force constant of the spring is 400 $N / m$ , by how much the spring will be compressed

1

2 \times 10^{- 2} m

2

0.2 m

3

20 m

4

200 m

Explanation:

K.E of block is converted into elastic P.E in spring i.e $\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

Work, Energy and Power

268662 A cradle is ' $h$ ' meters above the ground at the lowest position and ' $H$ ' meters when it is at the highest point. If ' $v$ ' is the maximum speed of the swing of total mass ' $m$ ' the relation between ' $h$ ' and ' $H$ ' is

1

(v^{2} / 2 g) + h = H

2

(v^{2} / g) + 2 h = H

3

(v^{2} / 2 g) + H = h

4

1 / 2 m v^{2} + h = H

Work, Energy and Power

268663 $A B$ is a frictionless inclined surface making an angle of $30^{\circ}$ with horizontal. A is $6.3 m$ above the ground while $B$ is $3.8 m$ above the ground. $A$ block slides down from $A$ , initially starting from rest. Its velocity on reaching $B$ is
![original image](https://cdn.mathpix.com/snip/images/DkkuvOboa_of-qrZXfk7iFbnlFFTxaIfbcZBLFuvRdQ.original.fullsize.png)

1

14.14 m / s

2

7.07 m / s

3

5 m / s

4

25 m / s

Explanation:

gain in $K . E =$ Loss of P.E $= m g (h_{1} - h_{2})$

Work, Energy and Power

268664 A stone of mass " $m$ " initially at rest and dropped from a height " $h$ " strikes the surface of the earth with a velocity " $v$ ". If the gravitational force acting on the stone is $W$ , then which of the following identities is correct

1

mv - mh = 0

2

1 / 2 {mv}^{2} - {Wh}^{2} = 0

3

1 / 2 m v^{2} - W h = 0

4

1 / 2 m v^{2} - m h = 0

Explanation:

gain in $K . E =$ Loss of P.E

Work, Energy and Power

268724 A block of mass $4 kg$ slides on a horizontal frictionless surface with a speed of $2 m / s$ . It is brought to rest in compressing a spring in its path. If the force constant of the spring is 400 $N / m$ , by how much the spring will be compressed

1

2 \times 10^{- 2} m

2

0.2 m

3

20 m

4

200 m

Explanation:

K.E of block is converted into elastic P.E in spring i.e $\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

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