**Q: 1. The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive of negative:**

**(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.**

**(b) Work done by gravitational force in the above case.**

**(c) Work done by friction on a body sliding down an inclined place.**

**(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.**

**(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.**

**Sol.**

**(a) It is positive, as force and displacement in same direction.**

**(b) It is negative, as the bucket is moving against the direction of gravitational force.**

**(c) It is also negative, as the friction is always opposite to the direction of motion.**

**(d) It is positive as the force and displacement are in same direction.**

**(e) It is negative as the resistive force acts against the direction of motion.**

**Q: 2. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Calculate the**

**(i) Work done by the applied force in 10s.**

**(ii) Work done by friction in 10s**

**(iii) Work done by the net force on the body in 10s**

**(iv) Change in K.E. of the body in 10s and interpret your result.**

**Sol. m = 2kg, u = 0, F = 7N, μ= 0.1, t = 10s, W = ?**

**Acceleration produced by applied force =**

**a _{1} = F/m**

**= 7/2 = 3.5 m/s ^{2}.**

**Force of friction = f = μR**

**= μmg = 0.1 × 2 × 9.8 = 1.96 N**

**Retardation produced by friction = a _{2} = (-f)/m**

**= (-1.96)/2**

**= 0.98 m/s ^{2}**

**Net acceleration =**

**a = a _{1} + a_{2}**

**= 3.50 – 0.98**

**= 2.52 m/s ^{2}**

**Distance moved by the body in 10 seconds = S = ut + (1/2) at ^{2}**

**= 0 + 1/2 × 2.52 × (10) ^{2}**

**= 126 m.**

**(i) Work done by applied force = W = F S**

**= 7 × 126 = 882 J**

**(ii) Work done by the frictional force**

**= f × s**

**= 1.96 × 126**

**= 246.9 J**

**(iii) Work done by the net force**

**= Net force × distance**

**= (7 – 1.96) × 126**

**= 635 J**

**(iv) Velocity at the end of 10 s is**

**v = u + at**

**= 0 + 2.52 × 10**

**= 25.2 m/s**

**Final k.E. = (1/2) mv ^{2} = 1/2 × 2 × (25.2)^{2} = 635 J**

**Initial K.E. = 0**

**Change in KE = 635 – 0**

**= 635 J**

**This shows that the change in K.E. is equal to the work done by the net force.**

**Q:3. The potential energy function for a particle executing simple harmonic simple harmonic motion is given by U(x) = (1/2)kx ^{2} , where K is force constant of the oscillator. For K = 0.5 N/M the graph of versus x is shown in the figure. Show that a particle of total energy 1 joule moving under this potential must turn back, when it reaches x = ± 2m.**

**Sol. At any instant, the total energy of an oscillator is equal to the sum of K.E. and P.E.**

**E = (1/2) mv ^{2} + (1/2) kx^{2}**

**When v = 0 ; E = (1/2) kx ^{2} At this instant the particle will comeback.**

**As E = 1 J and k = 1/2 N/m**

**1 = (1/2) × (1/2)x ^{2}**

**x = ±2m**

**Q: 4. A body is initially at rest. It undergoes one dimensional motion with constant acceleration. The power delivered to it at time t is proportional to**

**(i) t ^{1/2}**

**(ii) t**

**(iii) t ^{3/2}**

**(iv) t ^{2}**

**Sol. v = u + at = 0 + at = at**

**Power = P = F × v = ma × at = ma ^{2}t**

**∵ m and a are constant, p ∝ t**

**Q:5. A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to**

**(i) t ^{1/2}**

**(ii) t**

**(iii) t ^{3/2}**

**(iv) t ^{2}**

**Sol. P = F × v = [MLT ^{-2}] [LT^{-1}]**

**= [ML ^{2}T^{-3}] = constant**

**L ^{2}T^{-3} = constant**

**L ^{2}/T^{3} = constant**

**L ^{2} ∝ T^{3}**

**L ∝ T ^{3/2}**

**Displacement is proportional to t ^{3/2}**