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Work by gravity and spring force (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 "Work by gravity" and "Work by spring force".

The questions have been selected to enhance understanding of the topics covered in the modules titled " Work by gravity " and " Work by spring force " . 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 (Work by gravity and spring force)

Exercise 1

An object released from the top of a rough incline of height "h" reaches its bottom and stops there. If coefficient of kinetic friction between surfaces is 0.5, then what should be the work done by the external force to return the object along the incline to its initial position ?

(a) 1.5 mgh (b) 2mgh (c) 3.5 mgh (d) 4 mgh (a) 1.5 mgh (b) 2mgh (c) 3.5 mgh (d) 4 mgh

Solution

Speeds at initial and final poistions are same. Therefore, total work done by gravity and friction during downward motion is zero according to "work-kinetic energy" theorem. It means that work done during downward motion by gravity (mgh) equals work done by friction (-mgh), which is opposite in sign. During upward motion, both gravity and friction oppose motion of the object. Hence, work by external force must equal to the magnitude of negative works by gravity and friction.

W Ext = W Gravity + W Friction W Ext = 2 W Gravity W Ext = 2 m g h W Ext = W Gravity + W Friction W Ext = 2 W Gravity W Ext = 2 m g h

Note:
The work by external force is positive as direction of force and displacement are same.

Hence, option (b) is correct.

Exercise 2

Two springs, having spring constants k 1 k 1 and k 2 k 2 ( k 1 > k 2 k 1 > k 2 ), are stretched by same external force. If W 1 W 1 and W 2 W 2 be the work done by the external force on two springs, then :

(a) W 1 = k 1 k 2 W 2 (b) W 1 < W 2 (c) W 1 > W 2 (d) W 1 = W 2 (a) W 1 = k 1 k 2 W 2 (b) W 1 < W 2 (c) W 1 > W 2 (d) W 1 = W 2

Solution

Work done by the external force in stretching spring by "x" is :

W = 1 2 k x 2 W = 1 2 k x 2

We need to eliminate extension "x" in terms of "F" and "k" in order to make the comparison for same force. Now, external force on the spring is :

F = k x F = k x

Combining two equations, we have :

W = F 2 2 k W = F 2 2 k

For same external force,

W 1 W 2 = k 2 k 1 W 1 W 2 = k 2 k 1

As k 1 > k 2 k 1 > k 2 ,

W 1 < W 2 W 1 < W 2

Hence, option (b) is correct.

Exercise 3

A block is released from the height "h" along a rough incline. If "m" is the mass of the block, then speed of the object at the bottom of the incline is proportional to :

(a) m 0 (b) m 1 (c) m 2 (d) m 3 (a) m 0 (b) m 1 (c) m 2 (d) m 3

Solution

We need to obtain an expression for the speed "v". This relation can be obtained, using "work - kinetic energy" theorem :

Figure 1
Block on an incline
 Block on an incline  (wgsq1.gif)

1 2 m v 2 = m g l sin θ - μ k m g l cos θ 1 2 v 2 = g l sin θ - μ k g l cos θ 1 2 m v 2 = m g l sin θ - μ k m g l cos θ 1 2 v 2 = g l sin θ - μ k g l cos θ

Clearly, the speed at the bottom of the incline is independent of mass "m".

Hence, option (a) is correct.

Exercise 4

Two blocks of equal masses are attached to ends of a spring of spring constant "k". The whole arrangement is placed on a horizontal surface and the blocks are pulled horizontally by equal force to produce an extension of "x" in the spring. The work done by the spring on each block is :

Figure 2
Spring - blocks system
 Spring - blocks system  (wgsq2.gif)

(a) 1 2 k x 2 (b) 1 4 k x 2 (c) - 1 2 k x 2 (d) - 1 4 k x 2 (a) 1 2 k x 2 (b) 1 4 k x 2 (c) - 1 2 k x 2 (d) - 1 4 k x 2

Solution

The work done by a spring force for an extension "x" is :

W S = - 1 2 k x 2 W S = - 1 2 k x 2

However, spring does this work on blocks of same mass, which are pulled by equal horizontal force. The work by spring is, thus, equally divided on two blocks. Hence, work done by the sping on each block is :

W' S = - 1 4 k x 2 W' S = - 1 4 k x 2

Hence, option (d) is correct.

Exercise 5

An external force does 4 joule of work to extend it by 10 cm. What additional work (in Joule) is required to extend it further by 10 cm ?

(a) 4 (b) 8 (c) 12 (d) 16 (a) 4 (b) 8 (c) 12 (d) 16

Solution

The first part of the question can be used to determine the spring constant. The work done by external force is :

W = 1 2 k x 2 W = 1 2 k x 2

k = 2 W x 2 k = 2 x 4 0.1 2 = 800 N / m k = 2 W x 2 k = 2 x 4 0.1 2 = 800 N / m

Now, work done to extend the spring by a total of 20 cm,

W = 1 2 x 800 x 0.2 2 = 16 J W = 1 2 x 800 x 0.2 2 = 16 J

Thus, additional work required is 16 - 4 = 12 J

Hence, option (c) is correct.

Application level (Work by gravity and spring force)

Exercise 6

Rain drop falls from a certain height and attains terminal velocity before hitting the ground. If radius of rain drop is "r", then work done by the forces (gravity and viscous force) on the rain drop is proportional to :

(a) r (b) r 3 (c) r 5 (d) r 7 (a) r (b) r 3 (c) r 5 (d) r 7

Solution

Let "v" be the terminal speed. The change in kinetic energy ( 1 2 m v 2 1 2 m v 2 ) is equal to the work done by gravity and viscous force. Gravity does the positive work, whereas viscous force does negative force. When the drop attains terminal velocity, then viscous force equals the weight of the rain drop,

F = m g F = m g

6 π η r v = m g 6 π η r v = m g

v = m g 6 π η r v = 4 3 π r 3 ρ g 6 π η r v r 2 v = A r 2 v = m g 6 π η r v = 4 3 π r 3 ρ g 6 π η r v r 2 v = A r 2

where "A" is a constant. Now, work done by the forces :

W = 1 2 m v 2 W = 1 2 x 4 3 π r 3 ρ A 2 r 4 W r 7 W = 1 2 m v 2 W = 1 2 x 4 3 π r 3 ρ A 2 r 4 W r 7

Hence, option (d) is correct.

Answers


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

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