Work, Power and Energy

CHAPTER-4

WORK, POWER AND ENERGY

TRUE/FALSE

1. A simple pendulum with a bob of mass m swings with an angular amplitude of 40°. When its angular displacement is 20°, the tension in the string is greater than mg cos 20°. (1984; 2M)

positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The toal work done by the force F on the particle is : (1998; 2M) (a) – 2Ka2 _{(b) 2 Ka}2

(c) – Ka2 _{(d) Ka}2

7. A spring of force-sonstant k is cut into two pieces such that one piece is double of the other. Then the long piece will have a force-constant of :(1999; 2M) (a) (2/3)k (b) (3/2)k

(c) 3 k (d) 6 k

8. A wind-powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to:

(2000; 2M)

(a) v (b) v2

(c) v3 _{(d) v}4

9. A particle, which is constrained to move along x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F (x) = – k x + ax3_{. Here k and a are posiive} constant. For x ≥ 0, the functional form of the potential energy U (x) of the particle is : (2002; 2M)

U (X) X U (X) X (a) (b) U (X) X U (X) X (c) (d)

OBJECTIVE QUESTIONS

Only One option is correct :

1. Two masses of 1 g and 4 g are moving with equal kinetic energies. The ratio of the magnitudes of their

momenta is : (1980; 2M)

(a) 4 : 1 (b) _{2 : 1} (c) 1 : 2 (d) 1 : 16

2. A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to : (1984; 2M) (a) t1/2 _{(b) t}3/4

(c) t3/2 _{(d) t}2

3. A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is : (1985; 2M)

(a) MgL (b) MgL/3

(c) MgL/9 (d) MgL/18

4. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a_c is varying with time t as a_c = k2 _rt2_{, where k is a} constant. The power delivered to the particle by the force acting on it is : (1994; 1M) (a) 2π mk2 _r2 _{(b) mk}2_r2_t (c) 4 2 5 ( ) 3 mk r t (d) zero

5. A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is : (1998; 2M) (a) _u2_{– 2gL} (b) 2 gL

(c) _u2_–gL (d) 2

(

)

6. A force

_F

r

₌

₋

_K

₍

_y

_i

ˆ

₊

_x

_j

ˆ

₎

29 10. An ideal spring with spring-constnat k is hung from

the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in

the spring is : (2002; 2M) (a) 4 Mg k (b) 2 Mg k (c) Mg k (d) 2 Mg k

11. A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector _ar is correctly shown in (2002; 2M)

a → a → (a) (b)

→

_a

12. If W₁, W₂ and W₃ represent the work done in moving a particle from A to B along different paths, 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass m. Find the correct relation between W₁,

W₂ and W₃. (2003; 2M) B A 1 2 3 (a) W₁ > W₂ > W₃ (b) W₁ = W₂ = W₃ (c) W₁ < W₂ < W₃ (d) W₂ > W₁ > W3

13. A particle is placed at the origin and a force F = k x is acting on it (where k is a positive constant). If U (0) = 0, the graph of U (x) versus x will be (where U is the potential energy function : (2004; 2M)

U (x) x U (x) x (a) (b) U (x) x U (x) x (c) (d)

14. A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in : (2001; 2M)

V V

(a) (b)

V V

(c) (d)

15. A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A,

statisfies. (2008; 3M) θ v B L A (a) 4 π = θ (b) 2 4 π < θ < π (c) 4 2 π 3 < θ < π (d) π <θ<π 4 3

16. A block (B) is attached to two unstretched springs S₁ and S₂ with spring constants k and 4k , respectively. The other ends are attached to two supports M₁ and M₂ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small

distance x and released. The block returns and moves a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equilibrium

position of the block B. The ratio x y is (2008; 3M) 2 M2 _S 2 M1 S1 B 1 x 2 M2 _S 2 M1 S1 B 1 (a) 4 (b) 2 (c) 2 1 (d) 4 1

OBJECTIVE QUESTIONS

More than one options are correct?

1. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. the motion of the particle takes place in a plane. It follows that : (1987; 2M) (a) its velocity is constant

(b) its acceleration is constant (c) its kinetic energy is constant (d) it moves in a circular path

2. A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limits – φ and + φ. For an angular displacement θ (|θ| < φ), the tension in the string and the velocity of the bob are T and V respectively. The following relations hold good under the above conditions :

(1986; 2M) (a) T cos θ = Mg (b) T – Mg cos θ 2 MV L =

(c) The magnitude of the tangential acceleration of the bob |a_T| = g sin θ

(d) T = Mg cos θ

SUBJECTIVE QUESTIONS

1. In the figure (a) and (b) AC, DG and GF are fixed inclined planes, BC = EF = x and AB = DE = y. A small block of mass M is released from the point A. It slides down AC and reaches C with a speed V_C. The same block is released from rest from the point D. It slides down DGF and reaches the point F with speed V_F. The coefficients of kinetic frictions between the block and both the surface AC and DGF are µ. Calculate V_C

and V_F. (1980; 6M) A B C (a) D E F (b) G

2. A body of mass 2 kg is being dragged with a uniform velocity of 2 m/s on a rough horizontal plane. The coefficient of friction between the body and the surface is 0.20, J = 4.2 J/cal and g = 9.8 m/s2_{. Calculate the} amount of heat generated in 5 s. (1990; 5M) 3. A lead bullet just melts when stopped by an obstacle. Assuming that 25 percent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is 27°C

(Melting point of lead = 327°C, specific heat of lead = 0.03 cal/g-C°, latent heat of fusion of lead = 6 cal/g°C,

J = 4.2 J/cal). (1981; 3M)

4. Two blocks A and B are connected to each other by a string and a spring; the string passes over a frictionless pulley as shown in the figure. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of blocks is 0.2. Force constant of the spring is 1960 N/m. If mass of block A is 2 kg. Calculate the mass of block B and the energy stored in the spring. (1982; 5M)

B

A C

5. A 0.5 kg block slides from the point A (see Fig.) on a horizontal track with an initial speed of 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 N/m. The part AB of the track is frictionless and the part BC has the coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2 m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. (Take g = 10 m/s2₎ (1983; 7M)

31

A _{B D} C

3 m/s

6. A string, with one end fixed on a rigid wall, passing over a fixed fricionless pulley at a distance of 2 m from the wall, has a point mass M = 2 kg attached to it at a distance of 1 m from the wall. A mass m = 0.5 kg attached at the free end is held at rest so that the string is horizontal between the wall and the pulley and vertical beyond the pulley. What will be the speed with which the mass M will hit the wall when the mass

m is released? (1985; 6M)

m

M

7. A bullet of mass M is fired with a velocity 50 m/s at an angle θ with the horizontal. At the highest point of its trajectory, it collides head on with a bob of mass 3M suspended by a massless string of length 10/3 metres and gets embedded in the bob. After the collision the string moves through an angle of 120°. Find :

(i) the angle θ,

(ii) the vertical and horizontal coordinates of the initial position of the bob with respect to the point of firing of the bullet. (Take g = 10 m/s2₎

(1988; 6M) 8. A particle is suspended vertically

from a point O by an inextensible massles string of length L. A vertical line AB is at a distance L/ 8 from O as shown in figure. The object is given a horizontal velocity u.

O

L/8 A

L

u B

At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Find u. (1999; 10M)

9. A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see figure). The smaller sphere a has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less then d. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by θ. (2002; 5M) θ O R d Sphere B Sphere A

(a) Express the total normal reaction force exerted by the spheres on the ball as a function of angle θ. (b) Let N_A and N_B denote the magnitudes of the normal reaction forces on the ball exerted by the spheres A and B, respectively. Sketch the variations of N_A and N_B as function of cos θ in the range 0 ≤θ ≤ π by drawing two separate graphs in your answer book, taking cos θ on the horizontal axis. 10. A cart is moving along x-direction with a velocity of 4 m/s. A person on the cart throws a stone with a velocity of 6 m/s relative to himself. In the frame of reference of the cart the stone is thrown in y-z plane making an angle of 30° with vertical z-axis. At the highest point of its trajectory the stone hits an object of equal mass hung vertically from branch of a tree by means of a sring of length L. A completely inelastic collision occurs in which the stone gets embedded in the object. Determine : (1997; 5M) (i) the speed of the combined mass immediately after

the collision with respect to an observer on the ground.

(ii) the length L of the string such that tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass.

11. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m/s2_{, find the work done (in}

Joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

ASSERATION AND REASON

This question contains, statement I (assertion) and statement II (reason).

1. Statement-I : A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30° with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the meachnical energy in the second situation is smaller than that in the first situation. (2007; 3M)

Because :

Statement-II : The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination

(a) Statement-I is true, statement -II is true, statement-II is a correct explanation for statmeent-I (b) statemen-I is true, II is true;

statement-II is NOT a correct explanaion for statmeent-I (c) statement-I is true, statement-II is false (d) statement-I is false, statement-II is true

ANSWERS

TRUE/FALSE

1. T

OBJECTIVE QUESTION (ONLY ONE OPTION)

1. (c) 2. (c) 3. (d) 4. (b) 5. (d) 6. (c) 7. (b)

8. (c) 9. (d) 10. (b) 11. (c) 12. (b) 13.(a) 14. (a)15. (d)

16. (c)

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)

2. (b, c)

SUBJECTIVE QUESTIONS

1. V_C = V_F = 2(gy–µgx) 2. 9.33 cal 3. 409.8 m/s 4. 10 kg, 0.098 J

5. 4.24 m 6. 3.3 m/s

7. (i) θ = 30° (ii) The desired coordinates are (108.25m, 31.25m) 8. _       + = 2 3 3 2 gL u

9. (a) N = mg (3 cos θ – 2) (b) For θ ≤ cos–1 2

3    

 , NB 0, NA = mg (3cosθ – 2) and for θ ≥ cos –1 2 3      ; N_A = 0, N_B = mg (2 – 3 cos θ)

ASSERATION AND REASON

33 1. T - mg cos 20° 2 =mv R or T = mg cos 20° 2 +mv R 20° 20° T mg m mgcos20° mgsin20° ∴T > mg cos 20° (Qv≠0) ∴ (T)

OBJECTIVE QUESTIONS (ONLY ONE OPTION)

1. P = 2 Km or, 2 1 2 1 2 1 = = m m P P ∴ (c) 2.  =constant      = v dt dv m Fv dt vdv∝ ⇒ 2 3 2 1 t s t dt ds_{∝ ⇒ ∝} ∴ (c)

3. Considering reference for zero potential at the table surface,             − = 6 3 L g M Ui and U_f =0 ∴ 18 MgL U U W = _f − _i = ∴ (d) U = 0 L 3 2L/3 Before After 4. k rt v krt r v rt k a_c = ⇒ = 2 2⇒ = 2 2 2

Therefore, tangential acceleration, a_t= dv=kr dt Tangential force, F_t = ma_t = mkr Only tangential force does work. ∴ Power =F_t v = (mkr) (krt) or Power = mk2_r2_t

∴ (b)

5. From energy conservation v2 _{= u}2 _{– 2gL}

Now, since the two velocity vectors shown in figure are mutually perpendicular, hence the magnitude of the change of velocity will be given by

v u A B O ) ( 2 ) 2 ( | |∆vr = u2+v2 = u2+ u2− gL = u2−gL ∴ (d) 6. F =−K(yiˆ+xˆj) r

∫

∫

Force constant, k ∝ _{lengthof spring( )} 1 l ∴ k₁ = 3 2 k ∴ (b) 8. Power = . = ur r F v Fv F = _ _   dm v dt 2 ? ) ? ( Av Av v× = =

SOLUTIONS

TRUE/FALSE

where, ρ is density of wind A is area of cross-section of blades ∴ P = ρAv3 ⇒ 3 v P∝ ∴ (c) 9. dU = – F. dx ∴ U (x) = – 3 0 (– + )

∫

10. Let x be the maximum extension of the spring. From conservation of mechanical energy : decrease in gravitational potential energy = increase in elastic potential energy

M

M

11. When the bob is between mean and extream position it has both centripetal (ac)

r and tangential (at)

r

accelerations. The net acceleration (ar) is vector sum of ar_c and ar_t.

O at → a → ac → ∴ (c)

12. Gravitational field is a conservative force field. In a conservative force field work done is path independent. ∴ W₁ = W₂ = W₃ ∴ (b) 13. From F =–dU dx ( ) 0 0 0 – – ( ) = =

∫

∫

∫

14. Since, the block rises to the same heights in all the four cases, from conservation of energy, speed of the block at highest point will be same in all four cases. Say it is V₀. V0 N+mg Now,N + mg 2 0 mV R = or 2 0 _– mV N mg R =

R (the radius of curvature) in first case is minimum. Therefore, normal reaction N will be maximum in first case.

∴ (a)

Note : In the questions it should be mentioned that all the four tracks are frictionless. Otherwise V_o will be different in different tracks.

15. v= 5gL ...(i) gh v v 2 2 2 2 − =       ...(ii) ) cos 1 ( − θ = L h ...(iii)

Solving equations (i), (ii) and (iii), we get

8 7 cosθ=− _or = °      − = θ − ₁₅₁ 8 7 cos1 ∴ (d)

16. From energy conservation,

2 2 ) 4 ( 2 1 2 1 y k kx = ∴ _xy= ₂1 ∴ (c)

35 OBJECTIVE QUESTIONS (MORE THAN ONE OPTION) 1. The given case is of uniform circular motion, in which speed or kinetic energy is constant. Direction of velocity and acceleraion keep on changing although its magnitude remains constant.

∴ Correct options are (c) and (d).

2. From the FBD of bob,

T – mg cos θ L mV2 = and Mg sin θ = Ma_T or a_T = g sin θ ∴ (b) and (c) θ θ O T Mg sin θ Mg cos θ Mg L V M

SUBJECTIVE QUESTIONS

1. In both the cases work done by friction will be µ Mgx.

∴ 1 2 1 2 _–

2MVC = 2MVF =Mgy µMgx ∴ V_C = V_F = 2gy– 2µgx

2. S = vt = 2 × 5 = 10 m Q = work done against friction = µmgs = 0.2 × 2 × 9.8 × 10 = 9.33 cal

3. Heat energy required to just melt the bullet. Q = Q₁ + Q₂ Here, Q₁ = ms∆θ = (m × 103_{) (0.03 × 4.2) (327 – 27)} = (3.78 × 104 _{m) J} Q₂ = mL = (m × 103_{) (6 × 4.2) = (2.52 × 10}4_m) ∴ Q = (6.3 × 104_)m

Only 75% of kinetic energy is utilized to melt the bullet

0.75 × 1 2 2mv = Q

0.75 1 2

2

× × ×m v = 6.3 × 104_m ⇒ _{v = 409.8 m/s} 4. Normal reaction between blocks A and C will be zero.

Therefore, there will be no friction between them. Both A and B are moving with uniform speed. Therefore, net force on them shold be zero.

B T=kx f = µ m gB B m gA T=kx For equilibrium of A : m_Ag = kx ⇒ x = m gA k = (2)(9.8) 1960 = 0.01 m For equilibrium of B : µm_Bg = T = kx = m_Ag ∴ m_B = mA µ 2 0.2 = _{= 10 kg}

Energy stored in spring

U 1 2 2

= kx 1 2

= (1960) (0.01)2 _{= 0.098 J}

5. From A to B, there will be no loss of energy. Now let block compresses the spring by an amount x and comes momentarily to rest. By work-energy theorem,

kx f A B D _v=0 C x i f s f N mg W W W K K W + + + = − 2 2 2 1 2 1 ) ( 0 0+ −µmg BD+x − kx =− mv Substituting the values

(0.2) (0.5) (10) (2.14 + x) 1(0.5)(3) – (2)( )2 1 2

2 2

= x

Solving this equation, we get, x = 0.1 m

Now, spring exerts a force k x = 0.2N on the block. But to stop the block from moving limiting static friction is µ_s mg = (0.22) (0.5) (10) = 1.1 N. Since, 1.1 N > 0.2 N, block will not move further and it will permanenly stop there.

Therefore, total distance covered before it comes to rest permanently is

d = AB + BD + x = 2 + 2.14 + 0.1 = 4.24 m

6. Let M strikes with v. Then, velocity of m at this instant

will be v cos θ or 2

5v. Further M will fall a distance of 1 m while m will rise by

(

)

1 m 2 m m 1 m ( 5–1) m√ θ v cos θ √5 m θ v θ 1 2

∴ (2) (9.8) (1) = (0.5) (9.8)

(

)

7. (a) At the highest point, velocity of bullet is 50 cos θ. So, by conservation of linear momentum

M (50 cos θ) = 4M V_A ∴ V_A = 50 4       cos θ ...(1) At point, B, T = 0 but v ≠ 0 O V= θ VB 50 4 cos Hence, 4 Mg cos 60° = 3 50 2 ) 4 ( 2 2 = = ⇒V gl l V M B B (as 10 3 = l _{m and g = 10 m/s}2₎

By conservation of energy between A and B

      − = ⇒ − =V gh V V g l VB A B A 2 3 2 2 2 2 2 2 or, v2 _{= V}2 _{– 2g} 3 2      l or, cos? 100 4 50 3 50 100 2 2 2 _{ −}      =       ⇒ − = A B V V ∴ cos θ = 0.86 or θ = 30° (b) x = Range 2 2 1 sin2 2  _θ   =_{ } _    u g 50 50 3 2 10 2 × × = × × = 108.25 m y = H g u 2 ? sin2 2 = 50 50 1 2 10 4 × × = × × = 31.25 m Hence, the desired coordinates are (108.25 m, 31.25 m)

8. Let the string slacks at point Q as shown in figure. P to Q path is circular and beyond Q path is parabolic. At point C, velocity of particle becomes horizontal, therefore, QD = half the range of the projectile.

θ θ Q 90° – θ D L P u L cos θ L+L sin θ mg L 8 v

We have following conditions,

(a) T_Q = 0 Therefore, mg sin θ

2 = mv L ...(1) (2) v2 _{= u}2 _{– 2gh = u}2 _{– 2gL (1 + sin} θ_{) ...(2)} (3) QD 1 2 = (Range) ⇒       ₋ 8 ? cos L L 2 sin2(90 – ) 2 ° θ =v g 2 sin2 2 θ =v g       ₋ 8 1 ? cos ₌ 2         v gL sin θ cos θ ...(3) Substituting value of 2         v

gL = sin θ from Eq. (1) we get

      ₋ 8 1 ?

cos _{= sin}2θ_{. cos} θ _{= (1 – cos}2θ_{) cos}θ

or, cos? cos ? 8 1 ? cos _{− = −} 3 ∴ cos3?=₈1 or,cos θ = 1 2 or, θ = 60°

From Eq. (1) v2 _{= gL sin} θ _{= gL sin 60°} or v2 ₌ 3

2 gL

∴ Substituting this value of v2 _{in Eq. (2)} u2_=v2 _{+ 2gL (1 + sin} θ₎

37 = 3 2 1 3 2 2   + _ + _   gL gL =3 3 2 2 3 3 2 2   + = _ + _   gL gL gL u = 3 3 2 2   +       gL 9. (a) h (1–cos ) 2   =_ + _ θ   d R

Velocity of ball at angle θ

θ mg θ v h v2 _{= 2gh 2}

(

)

Let N be the total normal reaction (away from centre) at angle θ. Then, mg cos θ – N = 2 2  ₊      mv d R

Substituting value of v2 _{from Eq. (1) we get} mg cos θ – N = 2 mg (1 – cos θ) ∴ N = mg (3 cos θ – 2)

(b) The ball will lose contact with the inner sphere when N = 0

or, 3 cos θ – 2 = 0 or θ = cos–1 2

3      

After this it makes contact with outer sphere and normal reaction starts acting towards the centre. Thus for

θ ≤ cos–1 2 3       N_B = 0 and N_A = mg (3 cos θ – 2) and for θ ≥ cos–1 2

3      

N_A = 0 and N_B = mg (2 – 3 cos θ) The corresponding graphs are as follows

2/3 +1 cos θ mg NA –1 2/3 ₊₁cos θ 2mg 5mg NB

10. (i) Given Vr_cart =4iˆ m/s ∴ Vstone,cart=

r

(6 sin30°)j$ + (6 cos 30°)$k = (3j$ + 3 3 $k) m/s

∴ Vstone=Vstone,cart+Vcart =(4iˆ+3ˆj+3 3kˆ) r

r r

m/s At highest point of its trajectory, the vertical component (z) of its velocity will be zero, whereas the x and y-components will remain unchanged. Therefore, velocity of stone at highest point will be,

V r

= (4i$ + 3j$)m/s or speed at highest point,

V | V| r

= _{=( (4)}2₊₍₃₎2 _{m/s = 5m/s} Now, applying law of conservation of linear momentum, let V₀ be the velocity of combined mass after collision. Then, mV = (2m)V₀

∴ V₀ 5

2 2

V

= = m/s = 2.5 m/s

(ii) Tension in the string becomes zero at horizontal position. It implies that velocity of combined mass also becomes zero in horizontal position. Applying conservation of energy, we have

V0 = 2.5m/s L T = 0 V = 0 0 =V02– 2gL ∴ L 2 0 2 V g = 2 (2.5) 2(9.8) = = 0.32m ∴ Length of the string is 0.32 m.

11. Let the two blocks move with acceleration a and tension in the string be T.

3 10 10 36 . 0 72 . 0 36 . 0 72 . 0 _{× =} + − = a _m/s2_; 8 . 4 10 36 . 0 72 . 0 36 . 0 72 . 0 2 _{× =} + × × = T _N; m S 3 5 1 33 . 3 2 1_{× ×} 2 ₌ = 8 = T W J

ASSERATION AND REASON

1. In statement-I : Decrease in mechanical energ in case I will be

∆U₁= 1 2 2mv

But decrease in mechanical energy in case II will be

∆U₂= 1 2– 2mv mgh ∴ ∆U₂<∆U₁

or statement-I is correct.

In statement-II : Coefficient of fricition will not change or this statement is wrong.