Work Energy and Power Numerical

Question 1: 👇

A particle of mass approaches a region of force starting from r = +∞. The potential energy function in terms of distance from the origin is given by,

$U_{(r)}={K}/{2a^3}{(3a^2-r^2)}$ for 0 ≤ r ≤ a

$U_{(r)}={K}/{r}$ for r ≥ a where K>0 is a positive constant.

(a) Derive the force F(r) and determine whether it is repulsive or attractive.

(b) With what velocity should the particle start at r = ∞ to cross over to other side of the origin.

Solution:

(a) F_(r) = ? Nature of force whether it is repulsive or attractive =?

Concept: F= - {dU}/{dr}

$U_{(r)}={K}/{2a^3}{(3a^2-r^2)}$

Differentiating above equation wrt dr

For 0 ≤ r ≤ a $F= - {dU}/{dr} = - {d}/{dr} [{K}/{2a^3}{(3a^2-r^2)}]$

$F = - {d}/{dr} [{K}/{2a^3}{(0 - 2r)}]$

$F = [{K r}/{a^3}]$ F is positive.

For r ≥ a
$U_{(r)}={K}/{r }$

Differentiating above equation wrt dr

F= - {dU}/{dr} = - {d}/{dr} [{K}/{r}]

$F = {K}/{r^2}$ F is positive.

From above it is clear that F is positive in both cases so Force is repulsive in nature.

(b) at r = ∞ v = ? just to cross over to the other side of the origin.

A particle of mass approaches a region of force starting from r equals to plus infinity

From Law of conservation of energy,

(ME)₁ = (ME)₂

(ME)₁ = KE + PE

$KE = {1/2}{mv^2}$

PE =?

For r ≥ a $U_{(r)}={K}/{r }$

so, at r = ∞ U = 0

$ME_1 = KE + PE = {1/2}{mv^2}{+0} = {1/2}{mv^2}$

(ME)₂ = KE + PE

KE = 0 ( as velocity of particle should be equal to zero, just to pass other side of origin )

At origin PE = ?

For 0 ≤ r ≤ a $U_{(r)}={K}/{2a^3}{(3a^2-r^2)}$

at origin r = 0

$PE={K}/{2a^3}{(3a^2-0)} ={K}/{2a^3}{(3a^2)} ={3K}/{2a}$

$so, ME_2= {0+} {3K}/{2a}={3K}/{2a}$

(ME)_1 = (ME)_2

$so, {1/2}{mv^2} = {3K}/{2a}$

v = sqrt{(3K)/am}

Question 2: 👇

A single conservative force F(x) acts on a 1.0 kg particle that moves along the x-axis. The potential energy U_(x) is given by: U_(x)= 20 + (x–2)² where x is in meters. At x = 5.0 m the particle has a kinetic energy of 20 J.
(i) What is the mechanical energy of the system?
(ii) Make a plot of U_(x) as a function of x for –10 m < x < 10 m, and on the same graph draw the line that represents the mechanical energy of the system.
Use part (ii) to determine
(iii) The least value of x.
(iv) The greatest value of x between which the particle can move.
(v) The maximum kinetic energy of the particle.
(vi) The value of x at which it occurs.
(vii) Determine the equation for F (x) as a function of x.
(viii) For what value of x does F(x) = 0 ?

Solution:

(i) at x = 5m KE = 20J Find ME = ?

ME = KE + PE

To find ME, first of all we have to find PE, (U) = ?

U_(x) = 20+(x–2)²

at x = 5m U₍₅₎ = 20+(5–2)² = 29J

So, total mechanical energy (ME) = KE + PE = 20 + 29 = 49J

(ii) Plot U_(x) as a function of x, for -10m < x < 10m

A single conservative force Fx acts on a 1kg particle

Use part (ii) to solve next part,

(iii) The least value of x = ?

from graph x = -3.4m

(iv) The greatest value of x = ?

from graph x = 7.4m

(iv) Maximum KE = ?

For maximum kinetic energy , potential energy should be minimum (Law of conservation of energy)

from graph minimum value of potential energy = 20J

Total mechanical energy = 49J

So, maximum value of KE = ME – PE = 49J – 20J = 29J

(vi) KE is maximum and potential energy is minimum at x = 2m. { from par (v)}

(vii) F_(x) = ? F_(x) = – dU/dx

F_(x) = (2X – 4) N

(viii) When F_(x) = 0 x = ?

F_(x) = 2x – 4

0 = 2x – 4

x = 2m

You can see in graph also, slope of U-x graph (dU/dx) is zero at x = 2m

Question 3: 👇

A box having mass 400 kg is to be slowly slide through 10 m on a horizontal state track having friction coefficient 0.2 with the box. (a) Find the work done by the person pulling the box with a rope at an angle θ with the horizontal. (b) Find the work when the person has chosen a value of θ which ensures him the minimum magnitude of the force. (g = 10m/s²)

A box having mass 400 kg is to be slowly slide through 10 m on a horizontal state track having friction coefficient 0.2 with the box. Find the work done by ....

Solution: (a)

at equilibrium condition,

{R + F sin theta} = mg
{R = mg - F sin theta} ………………(1)

{F cos theta = f } ……………………(2)

and frictional force {f = mu R = mu( mg - F sin theta )}

putting value of f in equation (2)
{F cos theta = f = mu( mg - F sin theta )}
{F cos theta = mu mg - mu F sin theta }
{F cos theta + mu F sin theta = mu mg }
{F (cos theta + mu sin theta) = mu mg }
{F = mu mg / (cos theta + mu sin theta) }

Work done by the person pulling the box with a rope at an angle θ with the horizontal,

putting value of mu = 0.2, m = 400kg and g = 10ms^-2

{F = mu mg / (cos theta + mu sin theta) } = 0.2*400*10 / (cos theta + 0.2 sin theta) = 800 / (cos theta + 0.2 sin theta)

W = Fcos theta * d

W = 800 / (cos theta + 0.2 sin theta) {* cos theta * 10}

W = (8000cos theta) / (cos theta + 0.2 sin theta) = {(8000cos theta)*(1/{cos theta})} / {(cos theta + 0.2 sin theta) * (1/{cos theta})}

W = 8000 / ( 1 + 0.2 tan theta)

(b) minimum magnitude of the force = ?

Find dF/{d theta} = 0

Question 4: 👇

A block of mass m is pushed against a spring of spring constant k fixed at one end to a wall. The block can slide on a frictionless table as shown in the figure. The natural length of the spring is L₀ and it is compressed to one-fourth of the natural length and the block is released. Find its velocity as a function of its distance x from the wall and the maximum velocity of the block. The block is not attached to the spring.

Solution:

A block of mass m is pushed against a spring of spring constant k

After compression of spring to length ${L_0 /4}$ , spring is released, velocity of block will be different when (i) block is in contact with spring ${L_0 /4 < x < L_0}$ and (ii) contact break ${ x gt L_0}$ , block will move freely.
For ${L_0 /4 < x < L_0}$
From conservation of energy,
Mechanical energy when spring is compressed = Mechanical energy when spring is released
(KE)_1+(PE)_2 = (KE)_2 +(PE)_2

$0 + {1/2}{(3L_0 / 4)^2} = {1/2}{k (L_0 - x)^2} + {1/2}{mv^2}$

$v = sqrt{k/m ( (3L_0 / 4)^2 - (L_0 - x)^2)}$

For ${ x gt L_0}$
From conservation of energy,
Mechanical energy when spring is compressed = Mechanical energy when spring is released and block move freely = KE of block
(KE)_1+(PE)_2 = (KE)_2 +(PE)_2
$0 + {1/2}{(3L_0 / 4)^2} = {1/2}{mv^2}$

$v = 3L_0 / 4 sqrt{k/m}$

Question 5: 👇

A 1.2 kg collar C may slide without friction along a fixed smooth horizontal rod. It is attached to three springs each of constant K= 400 N/m and 150 mm un-deformed length. Knowing that, the collar is released from rest in the position shown in the figure.

Determine the maximum velocity it will reach in its motion. Here A, O and B are fixed points.

Question 6: 👇

A uniform string of mass ‘M’ and length 2a, is placed symmetrically over a smooth and small pulley and has particles of masses m and m’ attached to its ends; show that when the string runs off the peg its velocity is

Question 7: 👇

Particle 1 experiences a perfectly elastic collision with a stationary particle 2. Determine their mass ratio, if (a) after a head-on collision the particles fly apart in the opposite directions with equal velocities, (b) the particles fly apart symmetrically relative to the initial motion direction of particle 1 with the angle of divergence θ = 60°.

MY YouTube Channel Link : 👉🖱 https://www.youtube.com/channel/UCGpC7nWE0-bBv9I53MM8

A box having mass 400 kg is to be slowly slide through 10 m on a horizontal state track having friction coefficient 0.2 with the box. Find the work done by the person pulling the box with a rope at an angle θ with the horizontal. A box having mass 400 kg is to be slowly slide through 10 m on a horizontal state track having friction coefficient 0.2 with the box. Find the work done by the person pulling the box with a rope at an angle θ with the horizontal.

Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical, Work Energy and Power Numerical

Share the knowledge spread the love...

Work Energy and Power Numerical

Work Energy and Power Numerical

A particle of mass approaches a region of force starting from r = +∞. The potential energy function in terms of distance from the origin is given by,

Table of Contents

Leave a Reply Cancel reply