## A particle which is experiencing a force is given by F = 3 i – 12 j undergoes a displacement d = 4 i .

Q: A particle which is experiencing a force is given by $\vec{F} = 3 \hat{i} -12 \hat{j}$ undergoes a displacement $\vec{d} = 4 \hat{i}$ . If the particle had a kinetic energy of 3 J at the beginning of the displacement , what is its kinetic energy at the end of the displacement .

(a) 9 J

(b) 15 J

(c) 12 J

(d) 10 J

Solution : $\displaystyle W = \vec{F}.\vec{d}$

$\displaystyle W = (3 \hat{i} -12 \hat{j}).(4\hat{i})$

W = 12 J

Using Work energy Theorem ,

W = K2 – K1

12 = K2 – K1

K2 = 12 + K1

As initial kinetic energy = 3 J

K2 = 12 + 3

= 15 J

Correct option is (b)

## A force acts on a 2 kg object , so that its position is given as a function of time as x = 32 + 5 . What is the work done by this force in first 5 seconds .

Q: A force acts on a 2 kg object , so that its position is given as a function of time as x = 32 + 5 . What is the work done by this force in first 5 seconds .

(a) 850 J

(b) 900 J

(c) 950 J

(d) 875 J

Solution : x = 32 + 5

$\displaystyle \frac{dx}{dt} = 6 t$

v = 6 t

at t = 0 , v1 = 0

at t = 5 , v2 = 6 x 5 = 30 m/s

Work done = change in Kinetic Energy

$\displaystyle W = \frac{1}{2}m v_2^2 – \frac{1}{2}m v_1^2$

$\displaystyle W = \frac{1}{2}m (v_2^2 – v_1^2)$

$\displaystyle W = \frac{1}{2} \times 2 (30^2 – 0)$

W = 900 J

## A body of mass m = 10-2 kg is moving in a medium and experiences a frictional force F = -k v2 . Its initial speed is vo = 10 m/s ….

Q: A body of mass m = 10-2 kg is moving in a medium and experiences a frictional force F = -k v2 . Its initial speed is vo = 10 m/s . If after 10 sec its energy is $\frac{1}{8}m v_0^2$ , the value of k will be

(a) 10-3 kg/s

(b) 10-4 kg/m

(c) 10-1 kg/m-s

(d) 10-3 kg/m

Solution : Given F = -k v2

Acceleration $\displaystyle a = \frac{F}{m}$

$\displaystyle a = -\frac{k v^2}{m}$

$\displaystyle \frac{dv}{dt} = -\frac{k v^2}{m}$

$\displaystyle \frac{dv}{v^2} = -\frac{k}{m} dt$

$\displaystyle \int_{10}^{v} \frac{dv}{v^2} = -\frac{k}{m} \int_{0}^{t} dt$

$\displaystyle [-\frac{1}{v}]_{10}^{v} = -\frac{k}{m} t$

$\displaystyle [\frac{1}{v} – \frac{1}{10}] = \frac{k}{m} t$

$\displaystyle \frac{1}{v} = 0.1 + \frac{k}{m} t$

$\displaystyle v = \frac{1}{0.1 + \frac{k t}{m}}$

$\displaystyle v = \frac{1}{0.1 + \frac{k \times 10}{10^{-2}} }$

$\displaystyle v = \frac{1}{0.1 + 1000 k}$

$\displaystyle \frac{1}{2}m v^2 = \frac{1}{8} m v_o^2$

$\displaystyle v = \frac{v_o}{2} = \frac{10}{2}$

v = 5 m/s

$\displaystyle 5 = \frac{1}{0.1 + 1000 k}$

1 = 0.5 + 5000 k

k = 10-4 kg/m

Correct option is (b)

## A time dependent force F = 6 t acts on a particle of mass 1 kg . If the particle starts from rest , the work done by the force …

Q: A time dependent force F = 6 t acts on a particle of mass 1 kg . If the particle starts from rest , the work done by the force during the first 1 sec will be

(a) 22 J

(b) 9 J

(c) 18 J

(d) 4.5 J

Solution : Since , $\displaystyle \frac{dp}{dt} = F$

$\displaystyle dp = F dt$

$\displaystyle p = \int_{0}^{1} 6t dt$

$\displaystyle p = [3 t^2]_{0}^{1}$

p = 3 kg m/s

Kinetic Energy $\displaystyle K = \frac{p^2}{2m}$

$\displaystyle K = \frac{3^2}{2 \times 1}$

K = 4.5 J

Correct option is (d)

## A block of mass 2 kg is sliding on a smooth surface . At t= 0 its speed is v0 = 2 m/s . At t = 0 , a time varying force starts acting on the block …

Q: A block of mass 2 kg is sliding on a smooth surface . At t= 0 its speed is v0 = 2 m/s . At t = 0 , a time varying force starts acting on the block in the direction opposite to v0 . Find the speed of object (in m/s) at t= 10 sec

Solution : Area of F-t graph gives impulse

Impulse = change in momentum

-42 = m(v – v0)

-42 = 2(v – 2)

-21 = v -2

v = -19 m/s

## A particle of mass 1 kg is moving in the xy â€“ plane whose position at an instant t is r = 3 sin 2t i + 3(1 – cos2t)j , where r is in metre and t is in second…

Q: A particle of mass 1 kg is moving in the xy â€“ plane whose position at an instant t is $\vec{r} = 3 sin 2t \hat{i} + 3(1 – cos2t)\hat{j}$, where r is in metre and t is in second. The kinetic energy of the particle at instant t is

(a) 4.5 J

(b) 10 J

(c) 18 J

(d) Does not depend on the time

Solution: $\vec{r} = 3 sin 2t \hat{i} + 3(1 – cos2t)\hat{j}$

On differentiating w.r.t time

$\frac{d\vec{r}}{dt} = 6 cos 2t \hat{i} + 6 sin2t \hat{j}$

$\displaystyle \vec{v} = 6 cos 2t \hat{i} + 6 sin2t \hat{j}$

$\displaystyle v = \sqrt{(6 cos 2t)^2 + (6 sin2t)^2} = 6$

Kinetic Energy , $\displaystyle K = \frac{1}{2} m v^2$

$\displaystyle K = \frac{1}{2} \times 1 \times (6)^2$

K = 18 J

It is Independent of time .

Correct Options are (c) & (d)

## A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally …

Problem : Â A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed .Â  Find theÂ  speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the stringÂ  is equal to the weight of the particle.

Solution :

Let tension in the string become equal to the weight of the particle when particle reaches the point B and deflection of the string from vertical is q.Â  Resolving mg along the string and perpendicular to the string, we get net radial force on the particle at B i.e.

FR = T – mg cosÎ¸ Â  Â  Â  Â  Â  Â  Â  Â  Â  Â . . . (i)

If v be the speed of the particle at B, then

FR = Â  Â mv2/lÂ  Â  Â  Â  Â  Â  Â  Â  Â Â . . . (ii)

From (i)Â  and (ii),Â  we get

$\displaystyle T – m g cos\theta = \frac{m v^2}{l}$ Â Â  . . . (iii)

SinceÂ  at B,Â Â  T = mg

mg(1 â€“ cos Î¸) =Â mv2/l

v2 = gl (1 – cosÎ¸)Â  . . . . (iv)

Conserving the energy of the particle at point A and B,Â  we have

$\displaystyle \frac{1}{2}m v_0^2 = m gl(1-cos\theta) + \frac{1}{2}m v^2$

Where , $\displaystyle v_0 = \sqrt{g l} \; , and \; v = \sqrt{g l(1 – cos\theta)}$

gl = 2gl(1 – cosÎ¸) + gl (1 – cosÎ¸)

cosÎ¸ = 2/3Â Â Â Â Â Â Â Â Â Â Â Â  . . . . (v)

Putting the value of cos Î¸ in equation (iv) we get

$\displaystyle v = \sqrt{\frac{g l}{3}}$

## In the figure shown stiffness of the spring is k and mass of the block is m. The pulley is fixed.Â  Initially the block m is held such that…

Problem : Â  In the figure shown stiffness of the spring is k and mass of the block is m. The pulley is fixed.Â  Initially the block m is held such that, the elongation in the spring is zero and then released from rest. Find :

(a) the maximum elongation in the spring

(b) the maximum speed of the block m.

Neglect the mass of the spring, pulley and that of the string.

##### Solution:

(a)Â  Let the maximum elongation in the spring be x,Â  when the block is at position 2.

The displacement of the block mÂ  is also x.Â  IfÂ  E1Â  and E2Â  are the energies of the system when the block is at position 1 and 2 respectively. Then

E1 = U1g + U1s + T1
Where Â  U1g = gravitational P.E. with respect to surface S.

U1S = P.E. stored inÂ  the spring.

T1 = initial K.E. of the block.

E1 = mgh1 + 0 + 0 = mgh1 Â Â Â Â  . . . (1)

E2 = U2g + U2s + T2

$\displaystyle E_2 = mgh_2 + \frac{1}{2}kx^2 + 0$ Â Â Â Â Â Â Â Â Â Â Â  . . . (2)

From conservation of energy E1 = E2

$\displaystyle mgh_1 = mgh_2 + \frac{1}{2}kx^2$

$\displaystyle \frac{1}{2}kx^2 = m g (h_1 – h_2 )= mgx$

$\displaystyle x = \frac{2 m g}{k}$

(b) Â The speed of the block will be maximum when it is at the equilibrium point.Â  Let xo be the elongation in the spring when the block is at equilibrium point .

from conservation of energy

$\displaystyle m g h_1 = ( m g h_1 – x_0 ) + \frac{1}{2}mv^2 +\frac{1}{2}kx_0^2$

$\displaystyle mgx_0 = \frac{1}{2}mv^2 +\frac{1}{2}kx_0^2$

$\displaystyle mg \frac{mg}{k} = \frac{1}{2}mv^2 +\frac{1}{2}k (mg/k)^2$

$\displaystyle \frac{m^2 g^2}{2 k} = \frac{1}{2}mv^2$

$\displaystyle v = \sqrt{\frac{m}{k}}g$

## A wedge of mass M with a smooth quarter circular plane, is kept on a rough horizontal surface.Â …

ProblemÂ  : Â A wedge of mass M with a smooth quarter circular plane, is kept on a rough horizontal surface.Â  A particle of mass m is released from rest from the top of the wedge as shown in the figure.Â  When the particle slides along the quarter circular plane, it exerts a force on the wedge.Â  If the wedge begins to slide when the particle exerts a maximum horizontal force on it, find the coefficient of friction between the wedge & the horizontal surface.

Solution :Â

In order to find the coefficient of friction m at the time of maximum horizontal thrust (Fx)max exerted on the wedge by the particle given by the formula

Î¼Â = Â fmaxÂ /NÂ Â . . . (i)

Where fmax = maximum static friction between the wedge & ground.

(limiting friction ) that must be equal to the maximum horizontal force

(Fx)Â­max for prevalence of sliding of the wedge & N = normal force offered by the horizontal surface on the wedge.

â‡’ (F)max & N can be calculated as follows.Â  Let the particle attain a speed v at the angular position Î¸ as shown in the free body diagram. Since the particle accelerates towards the centre O with an acceleration a = v2/r , the force exerted on it must be radially inwards.

$\displaystyle R – mg sin\theta = \frac{m v^2}{r}$

$\displaystyle R = mg sin\theta + \frac{m v^2}{r}$ Â  Â . . .Â  (ii)

Conserving energy of the particle between position 1 & 2 we obtain

(Î”KE)1â†’2 = (Î”PE) 1â†’2

$\displaystyle \frac{1}{2}m v^2 = m g h$

$\displaystyle \frac{1}{2}m v^2 = m g (r sin\theta)$

$\displaystyle v = \sqrt{2 g r sin\theta}$ Â  . . . (iii)

Elimination of v between (ii) & (iii) yields

R = 3 mg sin Î¸ Â  Â  Â  Â  Â  Â  Â  Â  Â  Â . .. (iv)

The horizontal force acting on the wedgeÂ  = Fx = R cos Î¸

Fx =(3 mg sin Î¸) cos Î¸

Fx = 3 m g sin Î¸ cos Î¸

$\displaystyle F_x = \frac{3}{2}mg sin2\theta$ Â  .. . (v)

For Fx to be maximum sin 2Î¸ = 1

â‡’ Î¸Â = 450Â  Putting Î¸ in equation (v) horizontally

we obtain $\displaystyle F_{x(max)} = \frac{3}{2}mg$ Â  Â  Â  Â . . . (vi)

Resolving forces acting on the wedge for its equilibrium along horizontal & vertical we obtain,

fmax â€“ (Fx)max = Ma = 0

fmax = (Fx)max

fmax = (Fx)max =Â mg

N â€“ Mg â€“R sin Î¸ = May = 0

N = Mg + R sin Î¸

N = Mg + (3 mg sin Î¸ ) sin Î¸

$\displaystyle N = (M + \frac{3}{2}m) g$ (since Î¸ = Ï€/4)

Putting the values of fmax & N in equation (i) we obtain

$\displaystyle \mu = \frac{3 m}{2M + 3m}$

## A block of mass M initially has a velocity v0 when it just touches a spring. The block moves through a distance l before it stops …

Problem : Â  A block of mass M initially has a velocity v0 when it just touches a spring. The block moves through a distance l before it stops after compressing the spring. The spring constant is k and the coefficient of kinetic friction between block and table is m. As the block moves the distance I, (a) what is the work done on it by the spring force? Are there other forces acting on the block, and if so, what work do they do? (b) what is the total work done on the block? (c) use the work-energy theorem to find the value of l in terms of M, v0, m, g and k.

Solution:

(a) The net force acting on the block by the spring is equal to

$\displaystyle = \int \vec{F_{spring}}.\vec{ds}$

Work done by the spring

$\displaystyle = \int F_{spring} dx cos180$

$\displaystyle = -\int F_{spring} dx$

$\displaystyle = -\int_{0}^{l} k x dx$

$\displaystyle = – \frac{k l^2}{2}$

(b) The total work done , W = Î” KE

$\displaystyle = 0 – \frac{1}{2}m v_0^2$

$\displaystyle = – \frac{1}{2}m v_0^2$

(c) The work done by frictionÂ  = – Î¼ mgl

Total work done $\displaystyle = – \mu m g l – \frac{1}{2}k l^2$

$\displaystyle – \frac{1}{2}m v_0^2 = – \mu m g l – \frac{1}{2}k l^2$

$\displaystyle \frac{1}{2}m v_0^2 = \mu m g l + \frac{1}{2}k l^2$

$\displaystyle l = \frac{\mu m g}{k} [\sqrt{1+ \frac{k}{m}(\frac{v_0}{\mu g})^2}- 1]$