Work Energy and Power
7.0 Work energy theorem
7.0 Work energy theorem
Law of conservation of mechanical energy holds good only when conservative force acts on the system. What happen when non-conservative force acts on the system?
When both conservative and non-conservative force acts on the system work energy theorem holds good.
Work energy theorem states that the work done by all the forces (conservative and non-conservative) acting on a particle or system is equal to the change in kinetic energy.
Mathematically, $$\begin{equation} \begin{aligned} {W_{external}} + {W_{conservative}} + {W_{non - conservative}} + {W_{internal}} = K{E_{final}} - K{E_{initial}} \\ {W_{net}} = \Delta KE \\\end{aligned} \end{equation} $$
Note: We can apply work energy theorem in inertial as well as non-inertial frame of references.
Examples: A truck intially at rest $(u=0)$ started travelling with constant acceleration $a$ and acquired velocity $v$ after travelling a distance $s$. The friction between the floor of the truck and the block is sufficient to avoid sliding.
Now,
${v_b} = v = \sqrt {2as} $, velocity of block
${v_t}=v$, velocity of truck
${v_{bt}}={v_b}-{v_t}=0$, velocity of block with respect to the truck
Let us analyze and compare the above situation in inertial and non-inertial frame of refernece.
| Frame | Inertial frame of reference | Non-inertial frame of reference |
| FBD |   |   |
| Forces | $$f=ma$$ | $$\begin{equation} \begin{aligned} f = ma \\ {F_p} = ma \\\end{aligned} \end{equation} $$ |
| Work done | Net work done on the block for displacement $s$, $${W_{net}}=mas$$ | Net work done on the block for displacement $s$, $${W_{net}}=mas-mas$$ |
| Work energy theorem | Work done by all the forces $(W_{net})$ = change in kinetic energy $\left( {\Delta K{E_{inertial}}} \right)$ in an inertial frame of reference. $$\begin{equation} \begin{aligned} {W_{net}} = \Delta K{E_{inertial}} \\ mas = \frac{1}{2}m{\left( {{v_b}} \right)^2} \\ {v_b} = \sqrt {2as} \\\end{aligned} \end{equation} $$ | Work done by all the forces including pseudo force $(W_{net})$ = change in kinetic energy $\left( {\Delta K{E_{non-inertial}}} \right)$ in a non-inertial frame of reference. $$\begin{equation} \begin{aligned} {W_{net}} = \Delta K{E_{non - inertial}} \\ mas = \frac{1}{2}m{\left( {{v_{bt}}} \right)^2} \\ {v_{bt}} = 0 \\ \therefore {v_b} = {v_t} = \sqrt {2as} \\\end{aligned} \end{equation} $$ |
Note: Work done may be different in "intertial" and "non-inertial" frame of reference but the final conclusion will be same in both the cases.
Question 16. A rollercoaster cart of mass $m$ slides on a frictionless track, which ends as a straight horizontal section at point $B$. If the cart starts slipping from point $A$, how much far away from the track it will hit the ground.
Solution: As we know that the rollercoaster cart will gain kinetic energy by loosing its potential energy. So,
Initial potential energy, ${U_A}=mgH$
Final potential energy, ${U_B}=\frac{mgH}{2}$
Change in potential energy, $\Delta U=-\frac{mgH}{2}$
Work done, $W=-\Delta U=\frac{mgH}{2}$
By applying work energy theorem we get, $$\begin{equation} \begin{aligned} \sum W = \Delta KE \\ \frac{{mgH}}{2} = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2 \\\end{aligned} \end{equation} $$
${v_A}=0$, As the cart start from rest at point $A$. $$\begin{equation} \begin{aligned} \frac{{mgH}}{2} = \frac{1}{2}mv_B^2 \\ v_B^2 = gH \\ {v_B} = \sqrt {gH} \\\end{aligned} \end{equation} $$
It will follow projectile motion after leaving the track
Time after it will hit the ground, $$\begin{equation} \begin{aligned} {s_y} = {u_y}t + \frac{1}{2}{a_y}{t^2} \\ - \frac{H}{2} = \frac{1}{2}\left( { - g} \right){t^2} \\ t = \sqrt {\frac{H}{g}} \\\end{aligned} \end{equation} $$
Horizontal distance away from the track $(R)={v_B}t$, $$\begin{equation} \begin{aligned} R = \sqrt {gH} \times \sqrt {\frac{H}{g}} \\ R = H \\\end{aligned} \end{equation} $$
Question 17. Find the velocity of the block of mass $4m$ when it hits the wall.
Solution: First question is whether the block of mass $4m$ will fall down or not.
From the FBD of mass $4m$ we get, that the force is balanced in the horizontal direction but in the vertical direction there is no force to balance $4mg$. Therefore, the block will fall down.
Let $v_1$ & $v_2$ be the velocities of a block of mass $4m$ and $m$ respectively when the block of mass $4m$ hits the wall.
We choose the initial position of a block of mass $4m$ to be the reference plane.
Initial gravitational potential energy, $$\begin{equation} \begin{aligned} {U_i} = {\left( {{U_{4m}}} \right)_i} + {\left( {{U_m}} \right)_i} \\ {U_i} = 0 - mgx...(i) \\\end{aligned} \end{equation} $$
Final gravitational potential energy, $$\begin{equation} \begin{aligned} {U_f} = {\left( {{U_{4m}}} \right)_f} + {\left( {{U_m}} \right)_f} \\ {U_f} = - 4mg(1) - mg\left[ {x - \left( {\sqrt 5 - 1} \right)} \right]...(ii) \\\end{aligned} \end{equation} $$
From equations $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} \Delta U = {U_f} - {U_i} \\ \Delta U = \left( {\sqrt 5 - 5} \right)mg \\\end{aligned} \end{equation} $$
Work done by gravitation potential energy, $$\begin{equation} \begin{aligned} W = - \Delta U \\ W = \left( {5 - \sqrt 5 } \right)mg...(iii) \\\end{aligned} \end{equation} $$
Relation between $v_1$ and $v_2$ (By string constraint), $$\begin{equation} \begin{aligned} + {v_1}\cos \theta + 0 + 0 - {v_2} = 0 \\ {v_2} = {v_1}\cos \theta ...(iv) \\\end{aligned} \end{equation} $$
From work energy theorem we get, $$\begin{equation} \begin{aligned} W = \Delta KE \\ \left( {5 - \sqrt 5 } \right)mg = \frac{1}{2}4m{\left( {{v_1}} \right)^2} + \frac{1}{2}m{\left( {{v_2}} \right)^2} - (0 + 0)...(v) \\ \\\end{aligned} \end{equation} $$
From equation $(iv)$ and $(v)$ we get, $$\begin{equation} \begin{aligned} \left( {5 - \sqrt 5 } \right)mg = \frac{1}{2}4m{\left( {{v_1}} \right)^2} + \frac{1}{2}m{\left( {{v_1}\cos \theta } \right)^2} \\ As,\left( {\cos \theta = \frac{2}{{\sqrt 5 }}} \right) \\ \left( {5 - \sqrt 5 } \right)mg = \frac{1}{2}4m{\left( {{v_1}} \right)^2} + \frac{1}{2}m{\left( {{v_1}\frac{2}{{\sqrt 5 }}} \right)^2} \\ {v_1} = \sqrt {\frac{{5g\left( {5 - \sqrt 5 } \right)}}{{12}}} \\\end{aligned} \end{equation} $$
Question 18. How much the spring of constant $k$ be compressed so that the block of mass $m$ just reaches the top of the loop. The coefficient of friction is ${\mu _k}$ for horizontal track from point $A$ to $B$ as shown in the figure.
Solution: From FBD,$$N = mg...(i)$$
Frictional force $(f)$, $$f = {\mu _k}mg...(ii)$$
We have to find the compression in the spring, so that the block just recahes the top. Therefore, $${v_c}=0$$
Let the compression in the spring by $x$.
Applying work energy theorem we get, $$\begin{equation} \begin{aligned} \sum W = \Delta KE \\ {W_{spring}} + {W_{frictional}} + {W_{gravitational}} = K{E_f} - K{E_i} \\ \frac{1}{2}k{x^2} - {\mu _k}mgs - mg\left( {2R} \right) = 0 - 0 \\ \frac{1}{2}k{x^2} = {\mu _k}mgs + 2mgR \\ x = \sqrt {\frac{{4mgR + 2{\mu _k}mgs}}{k}} \\\end{aligned} \end{equation} $$
