State the conditions under which a force does not work. | Physics

WORK, POWER AND ENERGY

Very Short Question Answers

Q1. State the conditions under which a force does not work.
Ans.

If the displacement is zero(S=0).
if the force and displacement are perpendicular(F⊥S).
If the force is zero(F=0).

Q2. Define work, power and energy. State their S.I Units?
Ans.

Work: If a body is displaced from one place to another place by the application of the
external force, then the work is said to be done. S.I. Units: Joule.
Power: The rate of work done is known as power. S.I Units: Watt
Energy: It is the capacity to do work S.I Unit: Joule.

Q3. State the relation between kinetic energy and momentum of a body?
Ans.

$https://latex.codecogs.com/png.image?\inline \large \dpi{110} K= \frac{P^{2}}{2m}$

Q4. State the sign of work done by a force in the following a) Work done by a man 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
Ans.

Positive
Negative

Q5. State the sign of work done by a force in the following a) Work done by friction on a force in the following b) Work done by gravitational force in the above case.
Ans.

Negative
Positive

Q6. State the sign of work done by a force in the following a) Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.b) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Ans.

Positive
Negative

Q7. State if each of the following statements is true or false. Give reasons for your answer.a) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.b) The work done by earth’s gravitational force in keeping the moon in its orbit for its one revolution is zero.
Ans.

False. If external forces present, energy of the system changes.
True. Because gravitational force is conservative.

Q8. Which Physical Quantity Remains Constant.(i) In Elastic Collision (ii) in an Inelastic Collision
Ans.

elastic collision: Both momentum and kinetic energy
inelastic collision: momentum only

Q9. A body freely falling from a certain height ‘h’ after striking a smooth floor rebounds and rises to a height h . What is the coefficient of restitution between the floor and the body?
Ans.

$h_{1}=h,\: h_{2}=\frac{h}{2}\\ e=\sqrt{\frac{h_{2}}{h_{1}}}=\sqrt{\frac{\frac{h}{2}}{h}}=\sqrt{\frac{1}{2}}$

Q10. What is the total displacement of a freely falling body, after successive rebounds from the ground, before it comes to stop? Assume that ‘e’ is the coefficient of restitution between the body and the ground?
Ans.

The total displacement of a freely falling body, after successive rebounds from the
ground, before it comes to stop is ‘h’.

$$S=\frac{h(1+e^{2})}{1-e^{2}}$$

Short Question Answers

Q1. What is potential energy? Derive an expression for the gravitational potential energy?

Ans. WORK, POWER AND ENERGY

Q2. A lorry and a car moving with the same momentum are brought to rest by the application of brakes, which provide equal retarding forces. Which of them will come to rest in shorter time? Which will come to rest in less distance?

Ans. WORK, POWER AND ENERGY

Q3. Distinguish between conservative and non-conservative forces with one example each?

Ans. WORK, POWER AND ENERGY

Q4. Show that in the case of one dimensional elastic collision, the relative velocity of approach of two colliding bodies before collision is equal to the relative velocity of separation after collision?

Ans. WORK, POWER AND ENERGY

Q5. Show that two equal masses undergo oblique elastic collision will move at right angles after collision, if the second body initially at rest?

Ans. WORK, POWER AND ENERGY

Q6. Derive an expression for the height attained by a freely falling body after ‘n’ number of rebounds from the floor?

Ans. WORK, POWER AND ENERGY

Q7. Explain the law of conservation of energy?

Ans. WORK, POWER AND ENERGY

Essay Question Answers

Q1. Develop the notions of work and kinetic energy and show that it leads work-energy theorem?

Ans. Work:The scalar product of force and displacement is known as work.
$$ W=\vec{F}.\vec{S}$$

$$W=FS\cos\theta $$

Kinetic Energy: It is an energy possessed by a body by virtue of it's motion.
$$K=\frac12mv^{2}$$Work-EnergyTheorem:The work doneby netforce acting on abody is equal to change initskinetic energy.
$$W=K_f-K_i=\Delta K$$Proof:Consider abody of mass $m$ moving with an initial velocity $v_{0}$ and with a constant acceleration $a$. Suppose after covering a displacement $s$ the final velocity of the body $v$.we know that $$v^{2}-v_{o}^{2}=2as$$ multiplying with $\frac{1}{2}m$ on both sides

$$\begin{aligned}
&&\frac{1}{2}m(v^{2}-v_{o}^{2})=\frac{1}{2}m(2as) \\
&&\frac{1}{2}mv^{2}-\frac{1}{2}mv_{o}^{2}=mas \\
&&K_{f}-K_{i}=Fs \\
&&\therefore W=K_{f}-K_{i}=\Delta K \\
\end{aligned}$$

Q2. What are Collisions? Explain the possible the types of Collisions. Develop the theory of one dimensional elastic collisions?

Ans. Collision: A strong interaction between bodies involves exchange of momenta is called collision.

There are two types of collision:

Elastic Collision:The collision in which both momentum and kinetic energy are conserved are known as elastic collision.

Inelastic Collision: The collision in which kinetic energy is not conserved but momentum is conserved are known as Inelastic collision

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Derivation: As shown in diagram consider two bodies $m_1,m_2$ participated in one dimensional elastic collision with an initial velocities $\mathrm{u}_1,\mathrm{u}_2$ . Thus, after collision the final velocities of the bodies are $\mathrm{v}_{1},\mathrm{v}_{2}$ In the elastic collision $$\begin{array}{} {\text{According to the Law of Conservation of Linear Momentum}}\\ {m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2} \quad ----(1)}\\{m_{1}(u_{1}-v_{1})=m_{2}(v_{2}-u_{2}) \quad ----(2) }\\{\text{According to the Law of Conservation of Kinetic Energy}}\\{ \frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}=\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} }\\{m_{1}(u_{1}^{2}-v_{1}^{2})=m_{2}(v_{1}^{2}-u_{2}^{2}) \quad ----(3) }\\{\text{ Dividing eq(3) by eq(2)}}\\{\frac{m_{1}\left(u_{1}^{2}-v_{1}^{2}\right)}{m_{1}\left(u_{1}-v_{1}\right)}=\frac{m_{2}\left(v_{1}^{2}-u_{2}^{2}\right)}{m_{2}\left(v_{2}-u_{2}\right)} }\\{u_1+v_1=v_2+u_2 \quad ----(4)}\\{u_{1}-u_{2}=v_{2}-v_{1} }\end{array} $$Therefore, The relative velocity of approach before collision is equal to relative velocity of separation of after collision.
To find $V_{1}$ : from eq(4) $v_{2}=u_{1}-u_{2}+v_{1}$ is substituting in eq(1) $$v_{1}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)u_{1}+\left(\frac{2m_{1}}{m_{1}+m_{2}}\right)u_{2}$$To find $V_{2}$ : from eq(4) $v_{1}=u_{2}-u_{1}+v_{2}$ is substituting in eq(1)$$v_2=\left(\frac{2m_2}{m_1+m_2}\right)u_1+\left(\frac{m_2-m_1}{m_1+m_2}\right)u_2$$

Q3. State the law of conservation of energy and verify in the case of a freely falling body. What are the conditions under which the law of conservation of energy is applicable?

Ans. Law of conservation of Energy:Energy neither be created nor be destroyed.but it canbe change from one form to another form. The total mechanical energy of the system is remains constant.

Proof: As shown in diagram a body freely falling from a height $H$ above the ground..

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At point′A': velocity $V_{A}=0$ ,height $ h_{A} =H$, $a=g$

Kinetic Energy $K_{A}=\frac{1}{2}mV_{A}^{2}=0$

Potential Energy $U_{A}=mgh_{A}$ $$U_{A}=mgH $$

$\therefore$Total Energy$$\begin{aligned}E_{A}=K_{A}+U_{A}\\&E_{A}=0+mgH\\&E_{A}=mgH \quad-----(1)\end{aligned}$$

At point'B': velocity $V_{A}=0$ ,height $h_{B}=H-x$ $a=g$ velocity $v_{B}$ displacement $s=x$
$$\begin{aligned}
&V^{2}-V_{o}^{2}=2as \\
&V_{B}^{2}-0=2gx \\
&V_{B}^{2}=2gx
\end{aligned}$$

Kinetic Energy $K_{B}=\frac{1}{2}mV_{B}^{2}=\frac{1}{2}m(2gx)=mgx$

Potential Energy $U_{B}=mg(h-x)=mgH-mgx$

$\therefore$Total Energy $$\begin{aligned}E_{B}=K_{B}+U_{B} \\&E_{B}=mgx+mgH-mgx\\& E_{B}=mgH\quad-------(2)\end{aligned}$$

At point 'C': velocity $V_{A}=0$ ,$height h{C}=0$ $a=g$ velocity $V_{C}$ displacement $s=H$
$$\begin{aligned}
&V{C}^{2}-V_{A}^{2}=2as \\
&V_{C}^{2}-0=2gH \\
&V_{C}^{2}=2gH
\end{aligned}$$Kinetic Energy $K_{C}=\frac{1}{2}mv_{C}^{2}=\frac{1}{2}m(2gH)=mgH$

Potential Energy $U_{C}=mg(0)=0$

$\therefore$Total Energy $$\begin{aligned}E_{C}=K_{C}+U_{C}\\&E_{C}=mgH-0\\
&E_{C}=mgH \quad-------(3)\end{aligned}$$

From equations$ (1),(2) and (3)$ the total energy is equal at all three point. Hence the law of conservation of energy is verified.

Problems Question Answers

Q1. A test tube of mass 10 grams closed with a cork of mass 1 gram contains some ether. When the test tube is heated cork flies out under the pressure of the ether gas. The test tube is suspended horizontally by a weight less rigid bar of length 5cm. What is the minimum velocity with which the cork should fly out of the tube, so that test tube describing a full vertical circle about the point O. Neglect the mass of ether?

Ans. WORK, POWER AND ENERGY

Q2. A machine gun fires 360 bullets per minute and each bullet travels with a velocity of 600 m/s . If the mass of each bullet is 5gm, find the power of the machine gun?

Ans. WORK, POWER AND ENERGY

Q3. Find the useful power used in pumping 3425 m³ of water per hour from a well 8m deep to the surface, supposing 40% of the horse power during pumping is wasted.What is the horse power of the engine?

Ans. WORK, POWER AND ENERGY

Q4. A pump is required to lift 600kg of water per minute from a well 25m deep and to eject it with a speed of task 50m/s . Calculate the power required to perform the above?

Ans. WORK, POWER AND ENERGY

Q5. A block of mass 5kg initially at rest at the origin is acted on by a force along the X – Positive direction represented by F=(20 + 5x)N . Calculate the work done by the force during the displacement of the block from x = 0 to x = 4m.

Ans. WORK, POWER AND ENERGY

Q6. A block of mass 5kg is sliding down a smooth inclined plane as shown. The spring arranged near the bottom of the inclined plane has a force constant 600 N/m. Find the compression in the spring at the moment the velocity of the block is maximum?

Ans. WORK, POWER AND ENERGY

Q7. A force acts on a particle along the X-axis. Find the work done by the force in displacing the particle from x = +a to x = +2a. Take K as a positive constant?

Ans. WORK, POWER AND ENERGY

Q8. A force F acting on a particle varies with the position x as shown in the graph. Find the work done by the force in displacing the particle from x = -a to x =+2a?

Ans. WORK, POWER AND ENERGY

Q9. From a height 20m above a horizontal floor, a ball is thrown down with initial velocity 20m/s. After striking the floor, the ball bounces to the same height from which it was thrown. Find the coefficient of restitution for the collision between the ball the ball and the floor?

Ans. WORK, POWER AND ENERGY

Q10. A ball falls from a height of 10m on to a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is 1 . What is the total distance travelled by the before it ceases to rebound?

Ans. WORK, POWER AND ENERGY