KKMishra's Tutorials: Solutions to Problems on "Work and Energy" - H C Verma's Concepts of Physics, Part-I, Chapter-8, EXERCISES Q-11 TO Q-20

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WORK AND ENERGY:--
EXERCISES (11-20)

11. A box weighing 2000 N is to be slowly slid through 20 m on a straight track having friction coefficient 0.2 with the box, (a) Find the work done by the person pulling the box with a chain at an angle θ with the horizontal. (b) Find the work when the person has chosen a value of θ which ensures him the minimum magnitude of the force.

Answer: First let us draw a Free body diagram of the box to show the forces acting on it.

If the pull applied by the person is F,

Then Normal force N+F.sinθ=W = 2000 Newton

→N=2000-F.sinθ

Force of friction =µN =0.2x(2000-F.sinθ) = 400 -0.2xF.sinθ

Since the box is to be slowly slid, which means the driving force is just equal to frictional force.
This driving force = F.cosθ =400 - 0.2xF.sinθ
→F(cosθ+0.2sinθ)=400
→F(5cosθ+sinθ) = 400x5 (Multiply both sides by 5)
→F=2000/(5cosθ+sinθ)
Since the block is to be slid through a distance of 20 m
(a) So work done by the person =F.cosθx20 J
=2000xcosθx20/(5cosθ+sinθ) J
=40000/(5+tanθ) J

(b) For F to be minimum, dF/dθ = 0
So, -2000(-5sinθ+cosθ)/(5cosθ+sinθ)² = 0
→cosθ=5sinθ
→tanθ=1/5
(So F is minimum for that value of θ for which tanθ=1/5)
Putting this value in the work done expression we get,
40000/(5+1/5) =40000x5/26 = 7692 J

12. A block of weight 100 N is slowly slid up on a smooth incline of inclination 37° by a person. Calculate the work done by the person in moving the block through a distance of 2.0 m, if the driving force is (a) parallel to the incline and (b) in the horizontal direction.

Answer: (a) Driving force parallel to the incline

Since the block is slowly slid up, that means driving force parallel to incline is just equal to the resisting force which is a component of weight. See the figure below,

Figure for Q12

So, driving force F=100xsin37° N

Work done=Fxd =100xsin37°x2.0 =200x0.6 =120 J

(b) Driving force in the horizontal direction

When the driving force is in the horizontal direction, its component along incline will be equal to the force F= 100xsin37° N

So even in this case work done = 120 J

13. Find the average frictional force needed to stop a car weighing 500 kg in a distance of 25 m if the initial speed is 72 km/h.

Answer: Mass m=500 kg,
Initial speed u =72 km/h = 72000 m/3600 s = 20 m/s
Initial Kinetic Energy = ½mv² = ½x500 kgx (20 m/s)² = 250x400 J,
=1,00,000 J
Final speed v = 0, Final Kinetic Energy = 0
So work done by the average frictional force F = Change in Kinetic Energy =100000 - 0 = 100000 J
But also work done = Force x distance = Fxd (Distance d=25 m)
=Fx25 J
So, Fx25 = 100000
→F = 100000/25 = 4000 N

(Note: This problem can also be solved by calculating acceleration/retardation. Here 0²=20²-2ax25, a =400/50 =8 m/s²; Force=mass x acceleration =500x8 =4000 J)

14. Find the average force needed to accelerate a car weighing 500 kg from rest to 72 km/h in a distance of 25 m.

Answer: Mass m=500 kg,

Initial speed u = 0
Initial Kinetic Energy = 0
Final speed v =72 km/h =72000 m/3600 s = 20 m/s,
Final Kinetic Energy = ½mv² = ½x500 kgx (20 m/s)² = 250x400 J,
=1,00,000 J

So work done by the average force F = Change in Kinetic Energy =100000 - 0 = 100000 J

But also work done = Force x distance = Fxd (Distance d=25 m)
=Fx25 J
So, Fx25 = 100000
→F = 100000/25 = 4000 N

15. A particle of mass m moves on a straight line with its velocity varying with the distance traveled according to the equation v=a√x, where a is constant, Find the total work done by all the forces during a displacement from x=0 to x=d.

Answer: Initial velocity u (at x=0) =a√0 =0
Initial Kinetic energy =½mv² =0
Final velocity v (at x=d) =a√d
Final Kinetic energy = ½ma²d
Work done by all the forces during the displacement d = Change in kinetic energy = ½ma²d-0 =½ma²d.

16. A block of mass 2.0 kg kept at rest on an inclined plane of inclination 37° is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take g=10 m/s².

Answer: Let us draw a figure showing forces acting on the block

Figure for Q-16

Weight = mg =2x10 =20 N, Force F= 20 N

Net force along incline =F-mg.sinθ =20 - 20xsin37° =20-12.04 =7.96 N

Acceleration a = Force/mass = 7.96/2.0 =3.98 m/s²

Distance travelled = s = ut+½at² =0+0.5x3.98x1² =1.99 m

(a) Work done by the applied force = Force x distance = 20x1.99≈40 J

It is the maximum work done by the force because we have not considered the frictional force i.e. when the surface is smooth otherwise it will be even less.

(b) In the above situation the upward vertical movement of block during 1 s

=1.99xsin37° =1.20 m

So work done by the force of gravity (Downward) =mgx(-1.20) =-20x1.20 =-24 J

(c) Since the total work done on the block = Work done by the applied force + Work done by the gravity + Work done by the Normal force =40 J+ (-24 J) +0 =16 J

And Change in K.E = Final K.E.-Initial K.E. = Work done by all the forces = 16 J

So, Final K.E. =16 J (Since Initial K.E.=0)

17. A block of mass 2.0 kg is pushed down an inclined plane of inclination 37° with a force of 20N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of 10 m/s². If the block started from rest, find the work done (a) by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting on the block in the first second. Take g = 10 m/s².

Answer: First let us draw the free body diagram of the block, see figure below,

Figure for Q17

The forces on the block are, Weight W=mg =2x10 =20 N (Downward)

Applied force P =20 N (Down along the plane)

Normal force N =mg.cos37° =20x0.80 =16 N (up & perpendicular to the plane)

Frictional force F =µN (Up along the plane)

Now initial speed u=0, Acceleration a=10 m/s², Time t=1 s,

Final speed v=u+at =0+10x1 =10 m/s

And distance travelled =s =ut+½at² =0+0.5x10x1x1 =5 m

(a) So work done by the applied force in 1 s = Fxd =20 Nx5 m =100 J

(b) Since the weight acts downwards, distance travelled downward =5xSin37° m = 5x0.6 m =3.0 m

So work done by the gravity = 20 Nx3 m =60 J

Or in other way, component of gravity along the plane =mg.sin37° =20x0.6 N =12.0 N

Work done =12.0 Nx5 m =60 J

=½mv²-½mu² =0.5x2.0x10² -0 =100 J

Work done by the frictional force = Total W.D. by all the forces-W.D. by applied force-W.D. by the gravity-W.D. by the normal force

=100 J-100 J-60 J -0 (W.D. by the normal force is zero because movement in the direction perpendicular to plane is zero)

=-60 J

18. A 250 g block slides on a rough horizontal table. Find the work done by the frictional force in bringing the block to rest if is initially moving at a speed of 40 cm/s. If the frictional coefficient between the table and the block is 0.1, how far does the block move before coming to rest?

Answer: Mass m=250 g =0.25 kg
Initial speed u= 40 cm/s =0.40 m/s
Initial K.E. =½mu² =0.5x0.25x0.40x0.40 =0.02 J
Final K.E. = 0
Change in K.E. = 0.02 J =Work done by the frictional force

Frictional force = µmg =0.1x0.25x9.8 N =0.245 N
Let the block moves distance d before coming to rest,
so, Fxd=0.02
→d =0.02/0.245 =0.082 m =8.20 cm

19. Water falling from 50 m high fall is to be used for generating electric energy. If 1.8x10⁵kg of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100 W lamps can be lit?

Answer: Mass of water falling per second = 1.8x10⁵kg/3600 s = 50 kg/s
Potential energy of falling water per second = mgh =50 kgx9.8 m/s²x50 m =34500 J
So electric energy generated =½x24500 J/s =12250 W
Number of 100 W lamp that can be lit = 12250/100 =122.5
→122 (Removing the fractional part)

20. A person is painting his house walls. He stands on a ladder with a bucket containing paint in one hand and a brush in other. Suddenly the bucket slips from his hand and falls down on the floor. If the bucket with the paint had a mass of 6.0 kg and was at a height 2.0 m at the time it slipped, how much gravitational potential energy is lost together with the paint?

Answer: Mass of the paint with bucket = m =6.0 kg
Height h = 2.0 m
So potential energy of the paint with the bucket in the person's hand =mgh =6.0 kg x 9.8 m/s² x 2.0 m =117.6 J ≈118 J

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