We should here first think : why should potential energy change? A change in potential energy will take place when there is change in the water level or the level of ice mass. Now, ice cube is 90 % submerged in the water body.

Figure 1

Ice floating on water

When it melts, the volume of converted water is 90 % of the volume of ice. It means that level of water in the tank does not change. There is no change in potential energy, as far as the water body is concerned.

However, the level of ice body has changed. Its center was 5 cm below the top of the ice. When ice converts in to water, the center of the converted water body is 5.5 cm below. Thus, there is a change of level by 0.5 cm. The potential energy of the ice, therefore, decreases given by :

Figure 2

Change in potential energy
(a) (b)

Δ U = - m g h ⇒ Δ U = - V ρ g h ⇒ Δ U = - 0.1 3 x 0.9 x 10 x 0.5 = - 0.045 J Δ U = - m g h ⇒ Δ U = - V ρ g h ⇒ Δ U = - 0.1 3 x 0.9 x 10 x 0.5 = - 0.045 J

Hence, option (b) is correct.

Total mechanical energy of the particle,

E = U + K ⇒ K = E - U ⇒ K = E - 1 2 A x 2 ⇒ 1 2 m v 2 = E - 1 2 A x 2 ⇒ v = √ { 1 m ( 2 E - A x 2 ) } E = U + K ⇒ K = E - U ⇒ K = E - 1 2 A x 2 ⇒ 1 2 m v 2 = E - 1 2 A x 2 ⇒ v = √ { 1 m ( 2 E - A x 2 ) }

Hence, option (c) is correct.

The maximum horizontal range is :

R = v 2 g R = v 2 g

We can find out the speed of the ball using conservation of mechanical energy as :

K i + U i = K f + U f K i + U i = K f + U f

Here,

K i = 0 U i = 1 2 k x 2 K f = 1 2 m v 2 U f = 0 K i = 0 U i = 1 2 k x 2 K f = 1 2 m v 2 U f = 0

Thus,

1 2 k x 2 = 1 2 m v 2 v 2 = k x 2 m 1 2 k x 2 = 1 2 m v 2 v 2 = k x 2 m

and the maximum horizontal range is :

R = v 2 g = k x 2 m g R = 500 x 0.05 2 0.01 x 10 = 12.5 m R = v 2 g = k x 2 m g R = 500 x 0.05 2 0.01 x 10 = 12.5 m

Hence, option (c) is correct.

Let "v" be the speed of the block of 10 kg, when block of 5 kg is about to leave the ground. At that time :

k x = m g ⇒ x = m g k = 5 x 10 50 = 1 m k x = m g ⇒ x = m g k = 5 x 10 50 = 1 m

This problem suits the statement of conservation of energy stated as difference :

Δ U + Δ K = 0 Δ U + Δ K = 0

We note that there is change of level of the block of 5 kg only.

Δ U = - m g h = - 5 x 10 x 1 = - 50 J Δ U = - m g h = - 5 x 10 x 1 = - 50 J

and change in kinetic energy is also limited to block of 5 kg,

Δ K = 1 2 m v 2 = 1 2 x 10 x v 2 = 5 v 2 Δ K = 1 2 m v 2 = 1 2 x 10 x v 2 = 5 v 2

Putting in energy equation,

5 v 2 = 50 ⇒ v = √ ( 10 ) m / s 5 v 2 = 50 ⇒ v = √ ( 10 ) m / s

Hence, option (c) is correct.

The object attains velocity in vertical direction, when it reaches half of the total height. The acceleration of the object at any point in the motion is resultant of tangential and centripetal accelerations (provided by normal force) :

Figure 5

Circular motion in vertical plane

a = √ ( a T 2 + a R 2 ) a = √ ( a T 2 + a R 2 )

Here,

a T = g a T = g

and

a R = v 2 r a R = v 2 r

We need to know the speed of the object when it has reached half the height. Now, the minimum speed at the lowest point to enable the object to complete the revolution is :

v L = √ ( 5 g r ) v L = √ ( 5 g r )

From conservation of energy at lowest and middle levels,

K L + U L = K M + U M K L + U L = K M + U M

Considering zero reference potential at the lowest point,

U L = 0 U M = m g r U L = 0 U M = m g r

Putting these values in energy equation, we have :

1 2 m v 2 = 1 2 m { √ ( 5 g r ) } 2 - m g r ⇒ v 2 = { √ ( 5 g r ) } 2 - 2 g r = 3 g r 1 2 m v 2 = 1 2 m { √ ( 5 g r ) } 2 - m g r ⇒ v 2 = { √ ( 5 g r ) } 2 - 2 g r = 3 g r

and

a R = v 2 r ⇒ a R = 3 g r r = 3 g a R = v 2 r ⇒ a R = 3 g r r = 3 g

Thus,

⇒ a = √ { g 2 + ( 3 g ) 2 } = √ ( 10 ) g ⇒ a = √ { g 2 + ( 3 g ) 2 } = √ ( 10 ) g

Hence, option (d) is correct.

Here, we use the concept of work by an external force on a system. The system consists of two blocks and one spring. As friction is also involved, the corresponding energy statement is :

W Ext. = Δ U + Δ K + Δ E Thermal W Ext. = Δ U + Δ K + Δ E Thermal

W Ext. + W Friction = Δ U + Δ K W Ext. + W Friction = Δ U + Δ K

Since we are required to find the minimum force to initiate block on the left, we can consider that the system does not acquire kinetic energy during the process.

Δ K = 0 Δ K = 0

and

Δ U = 1 2 k x 2 Δ U = 1 2 k x 2

Let the right block moves by "x" before left block is initiated. The work by friction is done only on the right block :

W Friction = - μ m 2 g x W Friction = - μ m 2 g x

The work by external force is :

W Ext. = F x W Ext. = F x

Putting all these values in energy equation, we have :

⇒ F x - μ m 2 g x = 1 2 k x 2 ⇒ F x - μ m 2 g x = 1 2 k x 2

Also, for the condition that left block is just initiated,

k x = μ m 1 g k x = μ m 1 g

Combining two equations, we have :

⇒ F = μ g ( m 1 + m 2 2 ) ⇒ F = 0.5 x 10 x ( 1 + 2 2 ) = 10 N ⇒ F = μ g ( m 1 + m 2 2 ) ⇒ F = 0.5 x 10 x ( 1 + 2 2 ) = 10 N

Hence, option (a) is correct.