Q: The relationship between the force F and position x of a body is as shown in figure. The work done by force F , in displacing the body from x=1 \ m to x=5 \ m will be

Hint: Work done is the area under F-x graph. Correct Option: B \textbf{Step1: Calculation of work done} As work done = area enclosed by F-x graphWork done in displacing the body from x=1 m to x=5 m will be: W = Area of ABNM + Area of CDEN - Area of EFGH + Area of HIJ (Area of EFGH is taken negative as the force is in negative direction) \Rightarrow W = \left( {1 \times 10} \right) + \left( {1 \times 5} \right) - \left( {1 \times 5} \right) + \left( {\dfrac{1}{2} \times 10 \times 1} \right) = 15J So, the work done in displacing the body from x=1 m to x=5 m will be 15J .

Q: Two blocks of masses M_{1} and M_{2} are connected by spring of constant K . The spring is initially compressed and the system is released from rest at t = 0 second. The work done by spring on the blocks M_{1} and M_{2} be W_{1} and W_{2} respectively by time t. The speeds of both the blocks at time t are non zero. Then the value of \dfrac{W_{1}}{W_{2}} equals to

{ W }_{ 1 }=\cfrac { 1 }{ 2 } { M }_{ 1 }V_{ 1 }^{ 2 }\\ { W }_{ 2 }=\cfrac { 1 }{ 2 } { M }_{ 2 }{ V }_{ 2 }^{ 2 }\\ { M }_{ 1 }{ V }_{ 1 }={ M }_{ 2 }{ V }_{ 2 }\\ \cfrac { { W }_{ 1 } }{ { W }_{ 2 } } =\cfrac { { M }_{ 1 }{ V }_{ 1 } }{ { M }_{ 2 }{ V }_{ 2 } } .\cfrac { { V }_{ 1 } }{ { V }_{ 2 } } \\ \quad \quad \quad \quad =\cfrac { { M }_{ 2 } }{ { M }_{ 1 } }

Q: Two bodies move towards each other and collide inelastically. The velocity of the first body before impact is 2 m/s and of the second is 4 m/sec.The common velocity after collision is 1 m/s in the direction of the first body. How many times did the K.E. of the first body exceed that of the second body before collision.

Using momentum conservation m_{1} \times 2 - 4m_{2} = (m_{1} + m_{2}) \Rightarrow m_{1} = 5m_{2} Ratio of K.E =\dfrac{\dfrac{1}{2}\times m_{1} \times(2)^{2}}{\dfrac{1}{2}\times m_{2} \times(4)^{2}} =\dfrac{5m_{2}\times 4}{m_{2}\times 16} = 1.25

Question 1

A force  \vec{F}=(6\hat{i}-2\hat{j})N  acts on a particle that undergoes a displacement  \Delta \vec{r}=(3\hat{i}+\hat{j})m . Find (a) the work  done by the force on the particle and (b) the angle between  \vec{F}  and  \Delta\vec{r}

Accepted Answer

(a)  We use the mathematical representation of the definition of work:

Question 2

In each of the following, a force (see figure) F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Accepted Answer

Work done can be positive,negative or zero.Work done is positive when applied force and displacement is in same direction.Hence figure (b) shows positive work done.Work done is negative when applied force and displacement is in opposite direction.Hence figure (c) shows negative work done.Work done is zero when applied force and displacement is perpendicular.Hence figure (a) shows positive work done.

Question 3

A particle moves along the x-axis from  x=0  to  x=5  m under the influence of a force given by  F=7-2x+3x^2 . The work done in the process is

Accepted Answer

Work done is given by, \displaystyle W=\int_0^5 Fdx \displaystyle =\int_0^5(7-2x+3x^2)dx =7x-x^2+x^3 =135 \ J

Question 4

The relationship between the force  F  and position  x  of a body is as shown in figure. The work done by force  F , in displacing the body from  x=1 \ m  to  x=5 \ m  will be

Accepted Answer

&#10;&#10;Hint: Work done is the area under F-x graph.&#10;&#10;Correct Option: B 	extbf{Step1: Calculation of work done} As work done = area enclosed by F-x graphWork done in displacing the body from x=1 m to x=5 m will be:&#10;&#10;&#10;&#10;W = Area of  ABNM  + Area of  CDEN - Area of  EFGH  + Area of  HIJ &#10;&#10;(Area of  EFGH &#10;is taken negative as the force is in negative direction)&#10;&#10; &#10;\Rightarrow W = \left( {1 	imes 10} ight) + \left( {1 	imes 5} ight) -&#10;\left( {1 	imes 5} ight) + \left( {\dfrac{1}{2} 	imes 10 	imes 1} ight)&#10;= 15J &#10;&#10;So, the work done in displacing the body&#10;from x=1 m to x=5 m will be  15J  .

Question 5

A force  F = - K (y  \widehat i + x  \widehat j) , (where  K  is a +ve constant) acts on a particle moving in xy plane starting from origin, the particle is taken along the positive x-axis to the point ( a, 0 ) and then parallel to y axis to the point ( a, a ). The total work done by force  F  on the particle is

Accepted Answer

Work was done  \int Fdt When particle moves from  0  to  a  along x-axisThen  W_1 = \displaystyle \int_0^a - (Ky \widehat i) dx = 0           \because\,\, y=0  Now, when the particle moves parallel to the y-axis W_2 = \displaystyle \int_0^a - (Ka \widehat j) dy            \because\,\, x=a  	herefore W_2= - Ka \displaystyle \int_0^a dy = - Ka^2 	herefore W = W_1 + W_2 = - Ka^2

Question 6

A box of mass m is attached to a spring on a frictionless surface. The spring is compressed from its equilibrium position,  B , to point  A , a distance  x  from  B . Point  C  is also a distance  x  from  B , but in the opposite direction. When the box is released from position  A , Find out the points at which box's velocity is  maximum ?

Accepted Answer

When the box is released , it will do SHM ,because by Hook's law restoring force is directly proportional to displacement and opposite in direction now velocity  v  in SHM for displacement  x  is given by                      v=\omega\sqrt{{a}^{2}-{x}^{2}}     where  \omega  = angular frequency  and  a= amplitudefrom above relation it is clear that velocity   v  will be maximum at a point where  x  is minimum i.e.  x=0 and  x=0  shows the mean position  B .

Question 7

Two blocks of masses  M_{1}  and  M_{2}   are connected by spring of constant  K . The spring is initially compressed and the system is released from rest at  t = 0  second. The work done by spring on the blocks  M_{1}  and  M_{2}  be  W_{1}  and  W_{2}   respectively by time t. The speeds of both the blocks at time t are non zero. Then the value of  \dfrac{W_{1}}{W_{2}}  equals to

Accepted Answer

{ W }_{ 1 }=\cfrac {&#10;1 }{ 2 } { M }_{ 1 }V_{ 1 }^{ 2 }\ { W }_{ 2 }=\cfrac { 1 }{ 2 } { M }_{ 2 }{&#10;V }_{ 2 }^{ 2 }\ { M }_{ 1 }{ V }_{ 1 }={ M }_{ 2 }{ V }_{ 2 }\ \cfrac { { W&#10;}_{ 1 } }{ { W }_{ 2 } } =\cfrac { { M }_{ 1 }{ V }_{ 1 } }{ { M }_{ 2 }{ V }_{&#10;2 } } .\cfrac { { V }_{ 1 } }{ { V }_{ 2 } } \ \quad \quad \quad \quad =\cfrac&#10;{ { M }_{ 2 } }{ { M }_{ 1 } }

Question 8

Four blocks of masses  M_{1},M_{2},M_{3}  and  M_{4}  are placed on a smooth horizontal surface along a straight line as shown above. It is given that  M_{1}> > M_{2}> > M_{3}> > M_{4}.  All the blocks are initially at rest.  M_{1}  is given initial velocity  v_{0}  towards right such that it will collide with  M_{2}.  Consider all collisions to be perfectly elastic. The speed of  M_{4}  after all collision are over is

Accepted Answer

M_{1}  is very large as compared to  M_{2}.  Hence for collision between  M_{1}  and  M_{2}, M_{1}  can be considered equivalent to a wall and  M_{2}  as a small block. Thus the velocity of  M_{2}  will be  2v_{0}  after collision with  M_{1} . Similarly after collision between  M_{2}  and  M_{3} , the velocity of  M_{3}  will be   2(2v_{0}) . In sequence, the velocity of  M_{4}  shall be  2(2(2v_{0}))=8 v_{0}  after collision with  M_{3}.

Question 9

Two bodies move towards each other and collide inelastically. The velocity of the first body before impact is 2 m/s and of the second is 4 m/sec.The common velocity after collision is 1 m/s in the direction of the first body. How many times did the K.E. of the first body exceed that of the second body before collision.

Accepted Answer

Using momentum conservation m_{1} 	imes 2 - 4m_{2} = (m_{1} + m_{2}) \Rightarrow m_{1} = 5m_{2} Ratio of K.E =\dfrac{\dfrac{1}{2}	imes m_{1} 	imes(2)^{2}}{\dfrac{1}{2}	imes m_{2} 	imes(4)^{2}} =\dfrac{5m_{2}	imes 4}{m_{2}	imes 16} = 1.25

Question 10

A ball is dropped from a height  h  above a tile floor and rebounds to a height of  0.64h . The coefficient of restitution between the ball and the floor is

Accepted Answer

Let the velocity with which ball strikes the floor & rebounds be v and v' respectively.Using energy conservation: mgh=\dfrac { 1 }{ 2 } m{ v }^{ 2 }\quad \quad -(1)\ mg(0.64h)=\dfrac { 1 }{ 2 } m{ v' }^{ 2 }\quad \quad -(2) Using (1) & (2) \dfrac { 0.64 }{ 1 } =\dfrac { { v' }^{ 2 } }{ { v }^{ 2 } } \  \Rightarrow \dfrac { { v' } }{ { v } } =0.8 Therefore coefficient of restitution is  0.8

Question 11

A  2  kg body is dropped from height of  1  m onto a spring of spring constant  800  N/m as shown in the figure. A frictional force equivalent to  0.4  kgwt acts on the body. The speed of the body just before striking the spring will be (Take  g= 10\ m/s^2 )

Accepted Answer

We know that the change of kinetic energy is equal to work done by the system. Change of Kinetic energy  =\dfrac{1}{2}mv^2-0=\dfrac{1}{2}mv^2 Work done  =(mg-f)h  , where  f=0.4 kgwt =0. 4	imes 10=4N , friction force.Now,  \dfrac{1}{2}mv^2=(mg-f)h  \dfrac{1}{2}2v^2=(20-4)1 v^2=16 v=4\ m/s

Question 12

The block B and C in the figure have mass  m  each. The strings AB and BC are light, having tensions T _1  and T _2  respectively. The system is in equilibrium with a constant horizontal force  mg  acting on C. Then,

Accepted Answer

First from FBD of block C : T_2\sin	heta_2=F=mg .....(1) T_2\cos	heta_2=mg ....(2) (1)/(2), 	an	heta_2=1 \Rightarrow 	heta_2=45^o From (1),  T_2\sin45=mg \Rightarrow T_2=\sqrt{2}mg From FBD of block B: T_1\sin	heta_1=T_2\sin	heta_2 ....(3) T_1\cos	heta_1=mg+T_2\sin	heta_2 ...(4) (3)/(4), 	an	heta_1=\dfrac{T_2\sin	heta_2}{mg+T_2\sin	heta_2} put  	heta_2=45, T_2=\sqrt{2}mg ,  	an	heta_1=\dfrac{\sqrt 2mg(\dfrac{1}{\sqrt 2})}{mg(1+\sqrt 2.1/\sqrt 2)} or  	an	heta_1=\dfrac{1}{2} 	herefore \sin	heta_1=\dfrac{1}{\sqrt 5} now from (3), T_1=\dfrac{T_2\sin	heta_2}{\sin	heta_1}=\dfrac{\sqrt 2mg.\dfrac{1}{\sqrt 2}}{\dfrac{1}{\sqrt 5}} 	herefore T_1=\sqrt 5 mg

Question 13

A crane can lift up  10000\ kg  of coal in  1\ hour  from a mine of  180\ m  depth. If the efficiency of the crane is  80 \% , its input power must be:  [Take :  g = 10  \ ms^{-2} ]

Accepted Answer

Let input power  = P Since  80   % of this power is being used, so effective power is  \dfrac{80P}{100} Mass of coal  m = 10,000\ kg Time  t = 1 	imes 60 	imes 60\ s Height  h = 180\ m Gravitational acceleration  g = 10\ m/s^2  (given)Now, Power  =  work / timeSo,  \dfrac{80P}{100} = \dfrac{10,000 	imes 9.8 	imes 180}{60 	imes 60} Simplifying we get:  P = 6.25\ kW

Question 14

A boy blowing a whistle sends in air at  2\ gm/sec  with a speed of  150\ m/s . His lung power is

Accepted Answer

Power = \dfrac{ KE}{t}               = \dfrac{\dfrac{1}{2}m v^{2}}{t}               = \dfrac{1}{2}\dfrac{m}{t}v^{2}               = \dfrac{1}{2}	imes 2	imes 10^{-3}	imes (150)^{2}               = 22.5\ W

Question 15

A motor is used to deliver water at a certain rate through a given horizontal pipe. To deliver  n  times the water through the same pipe in the same time the power of the motor must be increased as follows.

Accepted Answer

Case I:Power  =\dfrac{\Delta K.E}{t}              =\dfrac{\dfrac{1}{2}mv^{2}}{t}=\dfrac{1}{2}dA.v^{3} where  d  is density  A  is area and  v  is velocity of waterCase II: P=\dfrac{1}{2}	imes dA.v^{3} When water delivered is n- times velocity will become n times i.e.  P  will become  n^{3}  times

Work, Energy And Power