Work, Energy, and Power

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(b) When drop falls first velocity increases, hence first KE also increases, after sometime speed is constant this is called terminal velocity, hence KE also become constant PE decreases continuously as the drop is falling continuously . the variation in PE and KE is best represented by b.

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(a) Mass m = 5kg

Radius =1m=R

Revolution per minute  w= 300rev/min

= 300 ( $2\pi$ )rad/min

= $\frac{\frac{300*2*3.14}{60}rad}{s}=\frac{10\pi rad}{s}$

Linear speed v= rw

= $\frac{300*2\pi }{60}=10\pi m/s$

KE= 1/2mv²

= $\frac{1}{2}*5*{10\pi }^2=250\pi$ ²J

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(c) When a pendulum oscillates in air, it will lose energy continuously in overcoming resistance due to air. Therefore total mechanical energy of the pendulum decreases continuously with time. This variation is correctly represented by option c.

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(d) When the earth is the closest to the sun, speed of the earth is maximum, hence KE is maximum. When the earth is the farthest from the sun speed is minimum hence, KE is minimum but never zero and negative.

This variation is correctly represented by option d

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(b) Power =constant

P+dw/dt=fds/dt

Fds=fdscos0

P=fds/dt=constant

Fv=constant

[MLT^-2] [LT^-1]= constant

[L²T^-3]= constant

L $\propto T$ ^3/2

This is a multiple choice answer as classified in NCERT Exemplar

(b) v=ax^3/2

m=0.5kg, a= 5m/s²

a= dv/dt= vdv/dt= ax^3/2d/dx (ax^3/2)

= ax^3/2a $*\frac{3}{2}*$ x^1/2= 3/2a²x²

Force = ma= m $\frac{3}{2}$ a²x²

Work done = ${\int }_0^{2}fdx={\int }_0^3\frac{3}{2}m{a}^2{x}^2$ dx

= $\frac{3}{2}$ ma² (x³/3)₀²

= $\frac{1}{2}m{a}^2*8=\frac{1}{2}*\left(0.5\right)*25*8$ =50J

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(b) When two bodies of equal masses collides elastically, their velocities are interchanged.

When ball 1 collides with ball 2 then velocity of ball1, v₁ becomes zero and velocity of ball 2, v₂ becomes v.

V₁=0, V₂=v

When ball 2 collides will ball 3 v₂ =0, v₃=v

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(b) Total energy is E=KE+PE……………1

When particle is at x =x_m i.e at extreme position, returns back. Hence at x=x_m and x=0 KE=0

From eqn1

E= PE+0 = PE= V (x_m)=1/2kx_m²

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(c) From conservation of mechanical energy

$\frac{1}{2}m{v}^2=mgh$

V= $\sqrt[]{2gh}$

Hence speed is same for both stones .

a₁=acceleration along inclined plane = gsin $\theta$ ₁

Similarly for stone a₂=gsin $\theta$ ₂ as $\theta$ ₂> $\theta$ ₁, hence a₂>a₁

This is a multiple choice answer as classified in NCERT Exemplar

(c) when we are considering the two bodies as system the total external force on the system will be zero . hence total linear momentum of the system remains conserved.