08: Work And Energy : Neet-xii-physics

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NEET-XII-Physics

08: Work and Energy

with Solutions - page 4

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Qstn #6
No work is done by a force on an object if
(a) the force is always perpendicular to its velocity
(b) the force is always perpendicular to its acceleration
(c) the object is stationary but the point of application of the force moves on the object
(d) the object moves in such a way that the point of application of the force remains fixed.
digAnsr: a,c,d
Ans : (a) the force is always perpendicular to its velocity
(c) the object is stationary but the point of application of the force moves on the object
(d) the object moves in such a way that the point of application of the force remains fixed.
No work is done by a force on an object if the force is always perpendicular to its velocity. Acceleration does not always provide the direction of motion, so we cannot say that no work is done by a force on an object if it is always perpendicular to the acceleration. Work done is zero when the displacement is zero.
In a circular motion, force provides the centripetal acceleration. The angle between this force and the displacement is 90`` °``, so work done by the force on an object is zero.
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Qstn #7
A particle of mass m is attached to a light string of length l, the other end of which is fixed. Initially the string is kept horizontal and the particle is given an upward velocity v. The particle is just able to complete a circle.
(a) The string becomes slack when the particle reaches its highest point.
(b) The velocity of the particle become zero at the highest point.
(c) The kinetic energy of the ball in initial position was
12mv2=mgl(d) The particle again passes through the initial position.
digAnsr: a,d
Ans : (a) The string becomes slack when the particle reaches its highest point.
(d) The particle again passes through the initial position.
The string becomes slack when the particle reaches its highest point. This is because at the highest point, the tension in the string is minimum. At this point, potential energy of the particle is maximum, while its kinetic energy is minimum. From the law of conservation of energy, we can see that the particle again passes through the initial position where its potential energy is minimum and its kinetic energy is maximum.
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Qstn #8
The kinetic energy force on the particle continuously increases with time.
(a) The resultant force on the particle must be parallel to the velocity at all instants.
(b) The resultant force on the particle must be at an angle less than 90° all the time.
(c) Its height above the ground level must continuously decrease.
(d) The magnitude of its linear momentum is increasing continuously.
digAnsr: b,d
Ans : (b) The resultant force on the particle must be at an angle less than 90° with the velocity all the time.
(d) The magnitude of its linear momentum is increasing continuously.
Kinetic energy of a particle is directly proportional to the square of its velocity. The resultant force on the particle must be at an angle less than 90° with the velocity all the time so that the velocity or kinetic energy of the particle keeps on increasing.
The kinetic energy is also directly proportional to the square of its momentum, therefore it continuously increases with the increase in momentum of the particle.
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Qstn #9
One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is
12kx2. The possible cases are
(a) at spring was initially compressed by a distance x and was finally in its natural length
(b) it was initially stretched by a distance x and and finally was in its natural length
(c) it was initially in its natural length and finally in a compressed position
(d) it was initially in its natural length and finally in a stretched position.
digAnsr: a,b
Ans : (a) at spring was initially compressed by a distance x and was finally in its natural length
(b) it was initially stretched by a distance x and and finally was in its natural length
For an elastic spring, the work done is equal to the negative of the change in its potential energy.
When the spring was initially compressed or stretched by a distance x, its potential energy is given by
`` {\left(P.E.\right)}_{i}=\frac{1}{2}k{x}^{2}``.
When it finally comes to its natural length, its potential energy is given by
`` {\left(P.E.\right)}_{f}=0``.
∴ Work done = `` -\left[{\left(P.E.\right)}_{f}-{\left(P.E.\right)}_{i}\right]=-\left[0-\frac{1}{2}k{x}^{2}\right]=\frac{1}{2}k{x}^{2}``
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Qstn #10
A block of mass M is hanging over a smooth and light pulley through a light string. T he other end of the string is pulled by a constant force F. The kinetic energy of the block increases by 20 J in 1 s.
(a) The tension in the string is Mg.
(b) The tension in the string is F.
(c) The work one by the tension on the block is 20 J in the above 1 s.
(d) The work done by the force of gravity is -20 J in the above 1 s.
digAnsr: b
Ans : (b) The tension in the string is F.
Tension in the string is equal to F, as tension on both sides of a frictionless and massless pulley is the same.
i.e., T - Mg = Ma
`` \Rightarrow ``T = Mg + Ma
So, the tension in the string cannot be equal to Mg.
The change in kinetic energy of the block is equal to the work done by gravity.
Hence, the work done by gravity is 20 J in 1 s, while the the work done by the tension force is zero.

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#
Section : iv

Qstn #1
The mass of a cyclist together with the bike is 90 kg. Calculate the increase in kinetic energy if the speed increases from 6⋅0 km/h to 12 km/h.
Ans : Total mass of the system (cyclist and bike), `` M={m}_{c}+{m}_{b}=90\,\mathrm{\,kg\,}``
Initial velocity of the system, `` u=6.0\,\mathrm{\,km\,}/\,\mathrm{\,h\,}=1.666\,\mathrm{\,m\,}/\,\mathrm{\,sec\,}``
Final velocity of the system, `` \nu =12\,\mathrm{\,km\,}/\,\mathrm{\,h\,}=3.333\,\mathrm{\,m\,}/\,\mathrm{\,sec\,}``
From work-energy theorem, we have:
`` \,\mathrm{\,Increase\,}\,\mathrm{\,in\,}\,\mathrm{\,K\,}.\,\mathrm{\,E\,}.=\frac{1}{2}M{\nu }^{2}-\frac{1}{2}m{u}^{2}``
`` =\frac{1}{2}90\times {\left(3.333\right)}^{2}-\frac{1}{2}\times 90\times {\left(1.66\right)}^{2}``
`` =499.4-124.6``
`` =374.8=375\,\mathrm{\,J\,}``
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Qstn #2
A block of mass 2.00 kg moving at a speed of 10.0 m/s accelerates at 3.00 m/s² for 5.00 s. Compute its final kinetic energy.
Ans : `` \,\mathrm{\,Mass\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,block\,},{M}_{b}=2\,\mathrm{\,kg\,}``
`` \,\mathrm{\,Initial\,}\,\mathrm{\,speed\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,block\,},u=10\,\mathrm{\,m\,}/s``
`` \,\mathrm{\,Also\,},a=3\,\mathrm{\,m\,}/{s}^{2}\,\mathrm{\,and\,}t=5\,\mathrm{\,s\,}``
`` \,\mathrm{\,Using\,}\,\mathrm{\,the\,}\,\mathrm{\,equation\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,motion\,},\,\mathrm{\,we\,}\,\mathrm{\,have\,}:``
`` \nu =u+at``
`` =10+3\times 5=25\,\mathrm{\,m\,}/s``
`` \therefore \,\mathrm{\,Final\,}\,\mathrm{\,K\,}.\,\mathrm{\,E\,}.=\frac{1}{2}m{\nu }^{2}``
`` =\frac{1}{2}\times 2\times 625=625\,\mathrm{\,J\,}``

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Qstn #3
A box is pushed through 4.0 m across a floor offering 100 N resistance. How much work is done by the resisting force?
Ans : Resisting force acting on the box, `` F=100\,\mathrm{\,N\,}``
Displacement of the box, S = 4 m
Also,`` \theta =180°``
`` ``
∴ Work done by the resisting force, `` ``
`` \,\mathrm{\,W\,}=\underset{\,\mathrm{\,F\,}}{\to }·\underset{\,\mathrm{\,S\,}}{\to }=100\times 4\times \,\mathrm{\,cos\,}180°=-400\,\mathrm{\,J\,}``
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Qstn #4
A block of mass 5.0 kg slides down an incline of inclination 30° and length 10 m. Find the work done by the force of gravity.
Ans : `` \,\mathrm{\,Mass\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,block\,},M=5\,\mathrm{\,kg\,}``
`` \,\mathrm{\,Angle\,}\,\mathrm{\,of\,}\,\mathrm{\,inclination\,},\theta =30°``
Gravitational force acting on the block, `` \,\mathrm{\,F\,}=mg``
Work done by the force of gravity depends only on the height of the object, not on the path length covered by the object.

`` \,\mathrm{\,Height\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,object\,},h=10\times \,\mathrm{\,sin\,}30°``
`` =10\times \frac{1}{2}=5\,\mathrm{\,m\,}``
`` \therefore \,\mathrm{\,Work\,}\,\mathrm{\,done\,}\,\mathrm{\,by\,}\,\mathrm{\,the\,}\,\mathrm{\,force\,}\,\mathrm{\,of\,}\,\mathrm{\,gravity\,},w=mgh``
`` =5\times 9.8\times 5=245\,\mathrm{\,J\,}``
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Qstn #5
A constant force of 2⋅5 N accelerates a stationary particle of mass 15 g through a displacement of 2⋅5 m. Find the work done and the average power delivered.
Ans : Given:
`` F=2.50\,\mathrm{\,N\,},S=2.5\,\mathrm{\,m\,}\,\mathrm{\,and\,}m=15g=0.015\,\mathrm{\,kg\,}``
Work done by the force,
`` W=F\mathit{·}S\,\mathrm{\,cos\,}0°\left(\,\mathrm{\,acting\,}\,\mathrm{\,along\,}\,\mathrm{\,the\,}\,\mathrm{\,same\,}\,\mathrm{\,line\,}\right)``
`` =2.5\times 2.5=6.25\,\mathrm{\,J\,}``
Acceleration of the particle is,
`` a=\frac{F}{m}=\frac{2.5}{0.015}``
`` =\frac{500}{3}\,\mathrm{\,m\,}/{\,\mathrm{\,s\,}}^{2}``
Applying the work-energy principle for finding the final velocity of the particle,
`` \frac{1}{2}m{v}^{2}-0=6.25``
`` \Rightarrow \nu =\sqrt{\frac{6.25\times 2}{0.15}}=28.86\,\mathrm{\,m\,}/s``
So, time taken by the particle to cover 2.5 m distance,
`` t=\frac{\nu -u}{\alpha }=\frac{\left(28.86\right)\times 3}{500}``
`` \therefore \,\mathrm{\,Average\,}\,\mathrm{\,power\,}=\frac{W}{t}``
`` =\frac{6.25\times 500}{\left(28.86\right)\times 3}=36.1\,\mathrm{\,W\,}``
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Qstn #6
A particle moves from a point
→r1=2 m →i+3 m →jto another point
→r2= 3 m →i+2 m →jacts on it. Find the work done by the force on the particle during the displacement.
Ans : Initial position vector,
`` \stackrel{\to }{{r}_{1}}=2\stackrel{\to }{i}+3\stackrel{\to }{j}``
Final position vector,
`` {\stackrel{\to }{r}}_{2}=3\stackrel{\to }{i}+2\stackrel{\to }{j}``
So, displacement vector,
`` \stackrel{\to }{r}={\stackrel{\to }{r}}_{2}-{\stackrel{\to }{r}}_{1}``
`` =\left(3\stackrel{\to }{i}+2\stackrel{\to }{j}\right)-\left(2\stackrel{\to }{i}+\stackrel{\to }{j}\right)``
`` =\stackrel{\to }{i}-\stackrel{\to }{j}``
`` \,\mathrm{\,Force\,}\,\mathrm{\,acting\,}\,\mathrm{\,on\,}\,\mathrm{\,the\,}\,\mathrm{\,particle\,},\stackrel{\mathit{\to }}{\mathit{F}}=5\stackrel{\to }{i}+5\stackrel{\to }{j}``
`` \,\mathrm{\,So\,},\,\mathrm{\,work\,}\,\mathrm{\,done\,}=\stackrel{\mathit{\to }}{\mathit{F}}\mathit{·}\stackrel{\mathit{\to }}{\mathit{S}}=5\times 1+5\left(-1\right)=0``
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Qstn #7
A man moves on a straight horizontal road with a block of mass 2 kg in his hand. If he covers a distance of 40 m with an acceleration of 0⋅5 m/s², find the work done by the man on the block during the motion.
Ans : Given:
`` \,\mathrm{\,Mass\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,block\,},m=2\,\mathrm{\,kg\,}``
`` \,\mathrm{\,Distance\,}\,\mathrm{\,coverd\,}\,\mathrm{\,by\,}\,\mathrm{\,the\,}\,\mathrm{\,man\,},s=40\,\mathrm{\,m\,}``
`` \,\mathrm{\,Acceleration\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,man\,},\,\mathrm{\,a\,}=0.5\,\mathrm{\,m\,}/{s}^{2}``
So, force applied by the man on the box,
`` F=ma``
`` =2\times \left(0.5\right)``
`` =1\,\mathrm{\,N\,}``
`` \,\mathrm{\,Work\,}\,\mathrm{\,done\,}\,\mathrm{\,by\,}\,\mathrm{\,the\,}\,\mathrm{\,man\,}\,\mathrm{\,on\,}\,\mathrm{\,the\,}\,\mathrm{\,block\,},W=F\mathit{·}S``
`` =1\times 40``
`` =40\,\mathrm{\,J\,}``
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Qstn #8
A force
F=α+bxacts on a particle in the x-direction, where a and b are constants. Find the work done by this force during a displacement from x = 0 to x = d.
Ans : Given that force is a function of displacement, i.e. `` \,\mathrm{\,F\,}=a+bx``,
where a and b are constants.
So, work done by this force during the displacement x = 0 to x = d,
`` \,\mathrm{\,W\,}=\underset{0}{\overset{d}{\int }}\,\mathrm{\,F\,}dx``
`` W=\underset{0}{\overset{d}{\int }}\left(a+bx\right)dx``
`` W={\left[ax+\frac{b{x}^{2}}{2}\right]}_{0}^{d}``
`` W=ad+\frac{b{d}^{2}}{2}``
`` \Rightarrow W=\left(a+\frac{bd}{2}\right)d``
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Qstn #9
A block of mass 250 g slides down an incline of inclination 37° with uniform speed. Find the work done against friction as the block slides through 1m.
Ans : `` \,\mathrm{\,Given\,}:``
`` m=250g,\,\mathrm{\,\theta \,}=37°,d=1\,\mathrm{\,m\,}``
Here, R is the normal reaction of the block.

As the block is moving with uniform speed,
`` f=mg\,\mathrm{\,sin\,}37°``
So, work done against the force of friction,
`` W=fd\,\mathrm{\,cos\,}0°``
`` W=(mg\,\mathrm{\,sin\,}37°)\times d``
`` W=(0.25\times 9.8\times \,\mathrm{\,sin\,}37°)\times 1.0``
`` W=1.5\,\mathrm{\,J\,}``
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