Physics Solutions

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)

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

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 v_o = 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)

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)