Work Power And Energy Practice Questions

We know force $(F)=k x$ $\;W=\;\frac{1}{2} k x^2$ $\;W=\;\frac{(k x)^2}{2 k}$ $\; \therefore W=\;\frac{F^2}{2 k}[\;\text{ as }\; F=k x]$When force is same then,$W \propto \;\frac{1}{k}$Given that, $W_1>W_2$ $\therefore k_1< k_2$ Statement-2 is true.For the same extension, ${\text{x}}_1= {\text{x}}_2= {\text{x}}$Work done on spring ${\text{S}}_1$ is ${\text{W}}_1=\;\frac{1}{2} k_1 x_1^2=\;\frac{1}{2} k_1 x^2$Work done on spring ${\text{S}}_2$ is ${\text{W}}_2=\;\frac{1}{2} k_2 x_2^2=\;\frac{1}{2} k_2 x^2$ $\therefore \;\frac{W_1}{W_2}=\;\frac{k_1}{k_2}$As $k_1< k_2$ then $W_1< W_2$So, Statement-1 is false.

Given $U(x)=\;\frac{a}{x^{12}}-\;\frac{b}{x^5}$ $U(x=\infty )=0$We know $F=-\;\frac{d U}{d x}=- [\;\frac{12 a}{x^{13}}+\;\frac{6 b}{x^7}]$At equilibrium: $\;\frac{d U(x)}{d x}=0$ $\;\Rightarrow \;\frac{-12 a}{x^{13}}=\;\frac{-6 b}{x^7}$ $\;\Rightarrow x= (\;\frac{2 a}{h}) ^{\;\frac{1}{6}}$ $\; \therefore U_{a t} \;\text{ equilibrium }\;=\;\frac{a}{ (\;\frac{2 a}{b}) ^2}-\;\frac{b}{ (\;\frac{2 a}{b}) }$ $\;=-\;\frac{b^2}{4 a}$ $\; \therefore D=0- (-\;\frac{b^2}{4 a}) =\;\frac{b^2}{4 a}$

The average speed of the athlete$v=\;\frac{100}{10}=10 m {/} s \;\; \therefore K . E .=\;\frac{1}{2} m v^2$If mass of athlete is $40 {\text{kg}}$ then, $K . E .=\;\frac{1}{2} \times 40 \times (10)^2=2000 J$If mass of athlete is $100 {\text{kg}}$ then, $K . E .=\;\frac{1}{2} \times 100 \times (10)^2=5000 J$His kinetic energy can be in the range $=2000 {\text{J}}$ to $5000 {\text{J}}$.

Let $u$ be the velocity with which the particle is thrown and $m$ be the mass of the particle. Then $K E=\;\frac{1}{2} m u^2$ .....(1) At the highest point the velocity is $u \cos 60^{\circ}$ (only the horizontal component remains, the vertical component being zero at the top-most point). Therefore kinetic energy at the highest point,$(K E)_H=\;\frac{1}{2} m u^2 \cos ^2 60^{\circ}$ $=\;\frac{K}{4}$ [From eq (1)]

Let the block compress the spring by $x$ before coming to rest. Initial kinetic energy of the block $=$ (potential energy of compressed spring) $+$ work done due to friction. $\;\frac{1}{2} \times 2 \times (4)^2=\;\frac{1}{2} \times 10000 \times x^{2+}15 \times x$ $10,000 x^{2+}30 x-32=0$ $\Rightarrow 5000 x^{2+}15 x-16=0$ $\therefore x=\;{-15 ± {\sqrt{(15)^{2-}4 \times (5000)(-16)}}}/{2 \times 5000}$ $=0.055 m=5.5 {\text{cm}}$

Kinetic energy at point of throwing is converted into potential energy of the particle during rise.$\;K . E=\;\frac{1}{2} m v^2=\;\frac{1}{2} \times 0.1 \times 25=1.25 J$ $\;W=-m g h=- (\;\frac{1}{2} m v^2) =-1.25 J$[As we know, $m g h=\;\frac{1}{2} m v^2$ by energy conservation ]

From work energy theorem we can say,Work done by tension + work done by force (applied) + Work done by gravitational force $=$ change in kinetic energy Here Work done by tension is zero$\;\Rightarrow 0+F \times A B-M g \times A C=0$ $\Rightarrow F=M g (\;\frac{A C}{A B}) =M g [\;{1-{1}/{{\sqrt{2}}}}/{\frac{1}{2}}]$as A B=ℓ sin \;45^{\circ}=\;{ℓ}/{{\sqrt{2}} and $A C=O C-O A=ℓ-ℓ \cos 45^{\circ}=ℓ (1-\;{1}/{{\sqrt{2}}})$where ℓ= length of the string. $\;\Rightarrow F=M g({\sqrt{2}}-1)$

Velocity is maximum when kinetic energy is maximum and when kinetic energy is maximum then potential energy should be minimumFor minimum potential energy, For minimum potential energy,$\;\;\frac{d V}{d x}=0$ $\;\Rightarrow x^{3-}x=0$ $\;\Rightarrow x=± 1$⇒ Min. Potential energy (P.E.) $;=\;\frac{1}{4}-\;\frac{1}{2}=-\;\frac{1}{4} J$ $\;K \cdot E \cdot (\max .)+P \cdot E \cdot (\min .)=2$ 𝜕(Given) $\; \therefore K \cdot E \cdot (\max .)=2+\;\frac{1}{4}=\;\frac{9}{4}$ $\; \therefore \;\frac{1}{2} m v_{\max }^2=\;\frac{9}{4}$ $\;\Rightarrow \;\frac{1}{2} \times 1 \times v^2{ }_{\max .}=\;\frac{9}{4}$ $\;\Rightarrow v_{\max }=\;{3}/{{\sqrt{2}}} \frac{\text{m}}{\ s}$

According to energy conservation law,Work done by the hand and due to gravity = total change in the kinetic energyInitially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy $\Delta K=0$ $W_{\;\text{hand }\;}+W_{\;\text{gravity }\;}=\Delta K$[Here distance covered would be $0.2$ meter for force by hand as force is applied while ball is in contact with hand.And gravity will still work while ball is in contact with hand so total distance due to gravity would be $2+0.2=2.2$ meter. $]$ $\Rightarrow F(0.2)-(0.2)(10)(2.2)=0 \Rightarrow F=22 N$ $\therefore$ Option (D) is correct.

Let $K$ be the initial kinetic energy and $F$ be the resistive force. Then according to work energy theorem,$W=\Delta K$ i.e., $3 F=\;\frac{1}{2} m v^{2-}\;\frac{1}{2} m (\;\frac{v}{2}) ^2$......(i) Let the bullet will penetrate ${\text{x}} {\text{cm}}$ more before coming to rest. $\therefore F x=\;\frac{1}{2} m (\;\frac{v}{2}) ^{2-}\;\frac{1}{2} m(0)^2$.......(ii) Dividing eq. (i) and (ii) we get,$\;\frac{x}{3}=\;\frac{1}{3}$ or $x=1 {\text{cm}}$

Loss in potential energy $=$ gain in kinetic energy$\;m \times g \times 80=\;\frac{1}{2} m v^2$ $\;\Rightarrow 10 \times 80=\;\frac{1}{2} v^2$ $\;\Rightarrow v^2=1600 \;\text{ or }\; v=40 \frac{\text{m}}{\ s}$

Let the length of the inclined plane is $=l$. So only $\;\frac{l}{2}$ part will have friction.According to work-energy theorem, $W=\Delta k=0$ (Since initial and final speeds are zero)$\therefore$ Work done by friction $+$ Work done by gravity $=0$i.e., $-(\mu m g \cos \varphi ) \;\frac{ℓ}{2}+m g ℓ sin \;\varphi =0$or $\;\frac{\mu }{2} \cos \varphi = sin \;\varphi$or $\mu =2 \tan \varphi$

$u=0 ; v=u+a T ; v=a T$Instantaneous power $=F \times v=m . a \cdot a t=m \cdot a^2 \cdot t$ $\therefore$ Instantaneous power $=\;\frac{m v^2 t}{T^2}$

Work done when the particle displaces from the origin, $W=\vec{F} \cdot \vec{x}$ $=(5 {i}^{∧}+3 {j}^{∧}+2 {k}^{∧}) \cdot (2 {i}^{∧}-{j}^{∧})$ $=10-3=7 {\text{J}}$

Mass of hanging part $(m^{'}) =\;\frac{4}{2} \times (0.6) k g=1.2 {\text{kg}}$Let at the surface $P E=0$Center of mass of hanging part $=0.3 {\text{m}}$ below the surface of the table$U_i=-m^{'} g x=-1.2 \times 10 \times 0.30=-3.6 {\text{J}}$ $\Delta U=m^{'} g x=3.6 {\text{J}}=$ Work done in putting the entire chain on the table.

Assume acceleration of body be $a$ $\therefore v_1=0+a T \Rightarrow a=\;\frac{v_1}{T}$ $\therefore v=a t \Rightarrow v=\;\frac{v_1 t}{T}$ $P_{\;\text{inst }\;}=\vec{F} \cdot \vec{v}=(m \vec{a}) \cdot \vec{v}$ $= (\;\frac{m v_1}{T}) (\;\frac{v_1 t}{T})$ $=m (\;\frac{v_1}{T}) ^2 t$

Given that, retardation $\propto$ displacement$\Rightarrow a=-k x$But we know $a=v \;\frac{d v}{d x}$ $\; \therefore \;\frac{v d v}{d x}=-k x$ $\;\Rightarrow ∫_{v_1}^{v_2} v d v=-k ∫_{0}^{x} x \text{dx}$ $\; (v_2^{2-}v_1^2) =-k \;\frac{x^2}{2}$ $\;\Rightarrow \;\frac{1}{2} m (v_2^{2-}v_1^2) =\;\frac{1}{2} m k (\;\frac{-x^2}{2})$ $\therefore$ Loss in kinetic energy is proportional to $x^2$.$\therefore \Delta K \propto x^2$

Work done by such force is always zero when a force of constant magnitude always at right angle to the velocity of a particle when the motion of the particle takes place in a plane.$\therefore$ From work-energy theorem, $\Delta K=0$ $\therefore K$ remains constant.