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.
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.
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
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
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..
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.