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# Mechanical energy (check your understanding)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Objective questions, contained in this module with hidden solutions, help improve understanding of the topics covered under the modules "potential energy", "Conservative force" and "Conservation of energy".

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The questions have been selected to enhance understanding of the topics covered under the modules "potential energy", "Conservative force" and "Conservation of energy". All questions are multiple choice questions with one or more correct answers. There are two sets of the questions. The “understanding level” questions seek to unravel the fundamental concepts involved, whereas “application level” are relatively difficult, which may interlink concepts from other topics.

Each of the questions is provided with solution. However, it is recommended that solutions may be seen only when your answers do not match with the ones given at the end of this module.

## Understanding level (Mechanical energy)

### Exercise 1

An ice cube of 10 cm floats in a partially filled water tank. What is the magnitude of change in gravitational potential energy (in Joule) when ice completely melts ? (Consider, relative density of ice = 0.9 and g = 10 m / s 2 m / s 2 )

(a) 0.04 (b) 0.045 (c) 0.05 (d) 0.055 (a) 0.04 (b) 0.045 (c) 0.05 (d) 0.055

#### Solution

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.

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 :

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.

### Exercise 2

The potential energy of a particle of mass "m" is given by 1 2 A x 2 1 2 A x 2 . If non-conservative force is not involved and total mechanical energy is "E", then speed of the particle at "x" is :

(a) ( E m - A x 2 ) (b) ( 2 E m - A x 2 ) (c) { 1 m ( 2 E - A x 2 ) } (d) { m ( 2 E - A x 2 ) } (a) ( E m - A x 2 ) (b) ( 2 E m - A x 2 ) (c) { 1 m ( 2 E - A x 2 ) } (d) { m ( 2 E - A x 2 ) }

#### Solution

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.

### Exercise 3

A ball of 10 gm is fired horizontally from a spring gun of spring constant 500 N/m. The spring is initially compressed by 5 cm and is released. The maximum horizontal distance (in meter) traveled by the ball is :

(a) 5 (b) 10 (c) 12.5 (d) 15 (a) 5 (b) 10 (c) 12.5 (d) 15

#### Solution

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.

### Exercise 4

The block of mass 10 kg is released is in the arrangement consisting of various elements as shown in the figure. If spring constant is 50 N/m, then the speed of 10 kg block, when block of 5 kg is about to leave the ground, is (Consider g = 10 m / s 2 m / s 2 ) :

(a) 1 (b) 2 (c) ( 10 ) (d) 10 (a) 1 (b) 2 (c) ( 10 ) (d) 10

#### Solution

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.

## Application level (Mechanical energy)

### Exercise 5

An object inside a spherical shell is projected horizontally at the lowest point such that it just completes the circular motion in a vertical plane. If the radius of the shell is 2 m, then acceleration of the object, when its velocity is vertical, is :

(a) 3 g (b) 5 g (c) 10 g (d) ( 10 ) g (a) 3 g (b) 5 g (c) 10 g (d) ( 10 ) g

#### Solution

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) :

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.

##### Note:
It is interesting to note that acceleration is independent of the radius of the spherical shell.

### Exercise 6

Two blocks of 1 kg and 2 kg are connected by a spring as shown in the figure. If spring constant is 500 N/m and coefficient of static and kinetic friction between surfaces are each 0.5, then what minimum constant horizontal force, F, (in Newton) is required to just initiate block on the left ? (Consider g = 10 m / s 2 m / s 2 ) .

(a) 10 (b) 20 (c) 30 (d) 40 (a) 10 (b) 20 (c) 30 (d) 40

#### Solution

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.

##### Note:
It is interesting to note that force required to initiate left block is independent of spring constant.


1. (b)    2. (c)    3. (c)    4. (c)
5. (d)    6. (a)


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