KKMishra's Tutorials: Solutions to Problems on "Work and Energy" - H C Verma's Concepts of Physics, Part-I, Chapter-8, EXERCISES Q-1 TO Q-10

My Channel on YouTube ---> SimplePhysics with KK

For links to

other chapters - See bottom of the page

WORK AND ENERGY:--
EXERCISES (1-10)

1. The mass of 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.

Answer: Initial speed = 6.0 km/h = 6000 m/3600 s = 5/3 m/s

K.E. = ½mv² =½*90*(5/3)² =45*25/9 =125 J

Final speed = 12 km/h =10/3 m/s

K.E. =½*90*(10/3)² =45*100/9 =500 J

So increase in K.E. =500 J-125 J =375 J

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,

Answer: Let us first compute velocity at the end of 5 s acceleration.

Here Initial velocity = u =10.0 m/s, acceleration a=3.00 m/s ² and time t=5.00 s.

Final velocity v=u+at =10+3*5 =25 m/s

So final K.E. =½mv² =½*2*25² =625 J.

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?

Answer: Resisting force F=100 N

Work done by this resisting force = 100 N*4 m =400 J

4. A block of mass 5.00 kg slides down an incline of inclination 30° and length 10 m. Find the work done by the force of gravity.

Answer: Force of gravity = mg =5*9.8 N =49 N (Downward)

Displacement along the direction of force of gravity

=10m*sin30° =5 m

So work done by the force of gravity =49 N*5 m =245 J

5. A constant force of 2.5 N accelerates a stationary particle of mass 15 g through a displacement of 2.50 m. Find the work done and the average power delivered.

Answer: Initial velocity u=0, Initial K.E. =0

Acceleration a=Force/mass =2.5/0.015 m/s² =2500/15 m/s²

=500/3 m/s² (Mass m=15/1000 =0.015 Kg)

Distance s =2.50 m

Final velocity v will be calculated using v²=u²+2as

→v² = 0+2*(500/3)*2.50 =1000*2.50/3 =2500/3

Final K.E. =½mv² =½*(15/1000)*2500/3 =5*25/20 =125/20 J

Work done = Change in K.E. =(125/20)-0 =6.25 J

Let us calculate the time t taken through this displacement.

From v=u+at

√(2500/3) =0+(500/3)t

→50/√3=500t/3

→t =150/500√3 s

Average Power = Work Done/time

=6.25/(150/500√3)

=6.25*500√3/150

=3125√3/150 =125√3/6 =125*1.73/6 =36.08 W

6. A particle moves from a point r₁= (2 m) i + (3 m) j to another point r₂= (3 m) i + (2 m) j during which a certain force F=(5 N) i + (5 N) j acts on it. Find the work done by the force on the particle during the displacement.

Answer: Displacement of the particle r = r₂-r₁

= (3 m) i + (2 m) j -{(2 m) i + (3 m) j}

= (1 m) i - (1 m) j

Work Done = Dot Product of vectors F and r.

= F.r

= {(5 N) i + (5 N) j}.{(1 m) i - (1 m) j}

= -(5 N*1 m) + (5N*1 m)

= -5 J + 5 J

= 0

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.

Answer: Velocity v after 40 m is given by, v²=u²+2as

(Where acceleration a = 0.5 m/s², Distance covered s=40 m)

v²=0+2*0.5*40 =40

Work done = Change in K.E.

=½mv²-½mu²

=½*2*40 -0

= 40 J

8. A force a+bx acts 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.

Answer: Displacement of the particle = d-0 = d

Force at x=0, F=a

and at x=d is F'=a+bd

Average force =(F+F')/2

= (a+a+bd)/2

= a+½bd

Work done = Force * Displacement

= (a+½bd)*d

9. A block of mass 250 g slides down an incline of inclination 37° with a uniform speed. Find the work done against the friction as the block slides through 1.0 m.

Answer: Mass m=250 g =0.25 kg

Weight =mg = 0.25*10 N =2.5 N (Taking g=10 m/s²)

The force against friction is the component of weight along the incline = mg.sin37°

(See figure below)

Displacement given = 1.0 m

So work done = Force x Displacement

= mg.sin37° x 1.0 J

= 2.5*sin37° J

= 2.5*0.60 J

= 1.50 J

10. A block of mass m is kept over another block of mass M and the system rests on a horizontal surface (Figure 8-E1). A constant horizontal force F acting on the lower block produces an acceleration F/2(m+M) in the system, the two blocks always move together. (a) Find the coefficient of kinetic friction between the bigger block and the horizontal surface. (b) Find the frictional force acting on the smaller block. (c) Find the work done by the force of friction on the smaller block by the bigger block during a displacement d of the system.