The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f. 2 volts and internal resistance 1 ohm. If a cell of e.m.f. 1.2 volts is balanced by it, the balancing length will be:
 
A)  90 cm
B)  60 cm
C)  50 cm
D)  75 cm​

Answer:

W = 1413.75 J

Explanation:

It is given that,

Mass of car, m = 650 kg

Initial speed of the car, u = 0.7 m/s

Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s

Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.

[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]

So, the work done by the car is 1413.75 J.

Answer:

a) h = 13,205.4 m

b)  r_f = 2.12 106 m

c)        e% = 0.68%

Explanation:

a) This is an exercise we are asked to use energy conservation,

Starting point. On the surface of Mimas

        Em₀ = K = ½ m v²

Final point. Where the ball stops

       [tex]Em_{f}[/tex] = U = m g h

        Em₀ = Em_{f}

        ½ m v² = m g h

         h = ½ v² / g

let's calculate

         h = ½ 41² / 0.0636

         h = 13,205.4 m

b) For this part we are asked to use the law of universal gravitation, write the energy

starting point. Satellite surface

           Em₀ = K + U = ½ m v² - GmM / r_o

final point. Where the ball stops

            [tex]Em_{f}[/tex]= U = - G mM / r_f

          Em₀ = Em_{f}

          ½ m v² - G m M / r_o = - G mM / r_f

In this case all distances are measured from the center of the satellite

         1 / rf = 1 / GM (-½ v² + G M / r_o)

let's calculate

         1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)

         1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)

          1 / r_f = 4,714 10⁻⁷

          r_f = 1 / 4,715 10⁻⁷

          r_f = 2.12 106 m

to measure this distance from the satellite surface

          r_f ’= r_f - r_o

          r_f ’= 2.12 106 - 1.98 105

         r_f ’= 1,922 106 m

c) the percentage difference is

          e% = 13 205.4 / 1,922 106 100

          e% = 0.68%

The estimate of part a is a little low

Answer:

a) The equation is [tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

b) Your velocity after collision is 2.64 m/s

c) The force you felt is 7392 N

d) you and your brother undergo an equal amount of acceleration

Explanation:

Your mass [tex]m_{y}[/tex] = 60 kg

your brother's mass [tex]m_{b}[/tex] = 30 kg

mass of the car [tex]m_{c}[/tex] = 80 kg

your initial speed [tex]u_{y}[/tex] = 0 m/s (since you've not started moving yet)

your brother's initial velocity [tex]u_{b}[/tex] = 3 m/s

your final speed [tex]v_{y}[/tex] after collision = ?

your brother's final speed [tex]v_{b}[/tex] after collision = ?

a) equations you need to use to figure out how fast you and your brother are moving after the collision is

[tex](m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

but [tex]u_{y}[/tex] = 0 m/s

the equation reduces to

[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

b) if your little brother reverses with velocity of 0.36 m/s it means

[tex]v_{b}[/tex] = -0.36 m/s (the reverse means it travels in the opposite direction)

then, imputing values into the equation, we'll have

[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

(30 + 80)3 = (60 + 80)[tex]v_{y}[/tex] + (30 + 80)(-0.36)

330 = 140[tex]v_{y}[/tex] - 39.6

369.6 = 140[tex]v_{y}[/tex]

[tex]v_{y}[/tex] = 369.6/140 = 2.64 m/s

This means you will also reverse with a velocity of 2.64 m/s

c) your initial momentum = 0  since you started from rest

your final momentum = (total mass) x (final velocity)

==>  (60 + 80) x 2.64 = 369.6 kg-m/s

If the collision lasted for 0.05 s,

then force exerted on you = (change in momentum) ÷ (time collision lasted)

force on you = ( 369.6 - 0) ÷ 0.05 = 7392 N

d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s

your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2

your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s

his deceleration is (3 - 0.36)/0.05 = 52.8 m/s

you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost

Answer:

35.3°

18.4°

Explanation:

a.

The first polariser polarises the unpolarised light reducing its intensity from I0 to I0/2. We have to reduce the intensity from I0/2 to I0/3.

Using to Law of Malus, I=I0cos²θ

cos²θ=I/I0=(I0/3)/I0/2 ,

cosθ=√2/3−−√=0.6667−−−−−√=0.8165

θ=cos−1(0.8165)=35.3∘

B.

Cos²θ=I/Io =Io/10/Io9

Cosθ= √9/10= 0.9487

= cos−10.9487

=18.4°

(a) The angle of polaroid such that intensity reduces by 1/3 is 35.26°

(b) The angle of polaroid such that intensity reduces by 1/10 is 63.43°

According to the Malus Law: The intensity of light when passing through a polarizer is given by:

I = I₀cos²θ

where θ is the angle of the polarizer axis with the direction of polarization of the light

I₀ is the initial intensity

When an unpolarised light passes through a polarizer, θ varies from 0 to 2π, so the intensity after passing the first polarizer is :

I = I₀<cos²θ>   { average of cos²θ, for 0<θ<2π}

I = I₀/2

Now, this emerging light passes through a second polarizer such that:

(a) the intensity is I' = I₀/3

From Malus Law:

I' = Icos²θ

I₀/3 =  (I₀/2)cos²θ

cos²θ = 2/3

θ = 35.26°

(b) the intensity is I' = I₀/10

From Malus Law:

I' = Icos²θ

I₀/10 =  (I₀/2)cos²θ

cos²θ = 1/5

θ = 63.43°

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Answer:

Acidosis is defined as the formation of excessive acid in the body due to kidney disease or kidney failure.

In order to compensate acidosis, the kidneys will reabsorb more HCO3 from the tubular fluid through tubular cells and collecting duct cell will secret more H+ and ammoniagenesis, which form more NH3 buffer.

Answer:

Amplitude, A = 5 m

Explanation:

Let a progressive wave is given by equation :

[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)

The general equation of a progressive wave is given by :

[tex]y=A\sin (\omega t-kx)[/tex] ....(2)

Here,

A is the amplitude of the wave

[tex]\omega[/tex] is the angular frequency

k is propagation constant

We need to find the amplitude of the wave.

If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.

Answer:

The velocity is  [tex]v = 2.84 1 \ m/s[/tex]

Explanation:

The  diagram showing this set up is shown on the first uploaded image (reference Physics website )

From the question we are told that

    The mass is  m =  4 kg

    The  length of the string is [tex]L = 2.0 \ m[/tex]

    The constant angle is  [tex]\theta = 35.4 ^o[/tex]

Generally the vertical forces acting on the mass to keep it at equilibrium vertically is mathematically represented as

           [tex]Tcos (\theta ) - mg = 0[/tex]

=>        [tex]mg = Tcos (\theta )[/tex]

Now let the force acting on mass horizontally be k  so from SOHCAHTOA rule

         [tex]sin (\theta ) = \frac{k }{T}[/tex]

=>      [tex]k = T sin \theta[/tex]

Now this k is also equivalent to the centripetal force acting on the mass which is mathematically represented as

          [tex]F_v = \frac{m v^2}{r}[/tex]

So

          [tex]k = F_v[/tex]

Which

=>       [tex]T sin \theta= \frac{ m v^2}{ r }[/tex]

So

        [tex]\frac{Tsin (\theta )}{Tcos (\theta )} = \frac{mg}{ \frac{mv^2}{r} }[/tex]

=>      [tex]Tan (\theta ) = \frac{v^2}{ r * g }[/tex]

=>      [tex]v = \sqrt{r * g * tan (\theta )}[/tex]

Now the radius is evaluated using SOHCAHTOA rule as

       [tex]sin (\theta) = \frac{ r}{L}[/tex]

=>    [tex]r = L sin (\theta)[/tex]

substituting values

       [tex]r = 2 sin ( 35.4 )[/tex]

       [tex]r = 1.1586 \ m[/tex]

So

       [tex]v = \sqrt{1.1586* 9.8 * tan (35.4 )}[/tex]

       [tex]v = 2.84 1 \ m/s[/tex]

Answer:

The intensity is [tex]I = 500 mW/m^2[/tex]

Explanation:

From the question we are told that

    The  intensity of the unpolarized light is [tex]I_o = 1000 \ m W /m^2 = 1000 *10^{-3} \ W/m^2[/tex]

Generally the intensity of the light emerging from the polarizer is  mathematically represented as

          [tex]I = \frac{I_o}{2}[/tex]

substituting values

         [tex]I = \frac{1000 *10^{-3}}{2}[/tex]

         [tex]I = 500 *10^{-3} W/m^2[/tex]

         [tex]I = 500 mW/m^2[/tex]

Answer:

32

Explanation:

Answer:

The  heat loss is  [tex]H = 8400\ W[/tex]

Explanation:

From the question we are told that

   The thickness is  [tex]t = 10 \ mm = 0.01 \ m[/tex]

    The inner temperature is  [tex]T_i = 25 ^oC[/tex]

    The outer temperature is [tex]T_o = 5 ^oC[/tex]

    The length of the window is  L  = 1 m  

    The  width of the window is  w  =  3 m  

Generally the heat loss is mathematically represented as

      [tex]H = \frac{k * A * \Delta T}{t}[/tex]

Where  k is the thermal conductivity of glass with value [tex]k = 1.4\ W/m \cdot K[/tex]

   and A  is the area of the window with value

           [tex]A = 1 * 3[/tex]

            [tex]A = 3 \ m^2[/tex]

substituting values

       [tex]H = \frac{1.4 * 3 * (23-5)}{0.01}[/tex]

       [tex]H = 8400\ W[/tex]

Answer:

the value of t = 0.49 seconds shows that its upward journey

and

at t = 5.43 seconds shows in downward journey

Explanation:

Given:

initial speed, u = 29 m/s

acceleration due to gravity, g = - 9.8 m/s^2

h = 13 m

Let it is moving with velocity v at a height of 13 m.

Use third equation of motion

v² = u² + 2gh

By substituting the values

v² = 29² - (2 * 9.8 * 13)

v = sqrt 585.94

v  = 24.2 m/s

Let it takes time t to reach at height 13 m

Use second equation of motion

s = u * t + 1/2 * g * t²

13 = 29t - 4.9t²

4.9t² - 29t + 13 = 0

using quadratic equation to solve time

     29 ± [tex]\sqrt{29^2 - 4 * 4.9 * 13}\\[/tex]

t = ------------------------------------

                     2 * 4.9

t = 5.43 second or t = 0.49 second

Therefore...

the value of t = 0.49 seconds shows that its upward journey

and

at t = 5.43 seconds shows in downward journey

Answer:

118 minutes( 2 hours approximately )

Explanation:

Here, we are interested in calculating the orbital period of the satellite

Please check attachment for complete solution

Answer:

T = 7101 s = 118.35 mins = 1.9725 hrs

Explanation:

To solve the question, we apply the formula for gravitational acceleration

a = GM/r², where

a = acceleration due to gravity

G = gravitational constant

M = mass of the earth

r = distance between the satellite and center of the earth

Now, if we make r, subject of formula, we have

r = √(GM/a)

Recall also, that

a = v²/r, making v subject of formula

v = √ar

If we substitute the equation of r into it, we have

v =√a * √r

v =√a * √[√(GM/a)]

v = (GM/a)^¼

Again, remember that period,

T = 2πr/v, we already have v and r, allow have to do is substitute them in

T = 2π * √(GM/a) * [1 / (GM/a)^¼]

T = 2π * (GM/a³)^¼

T = 2 * 3.142 * [(6.67*10^-11 * 5.97*10^24) / (6.25³)]^¼

T = 6.284 * [(3.982*10^14) / 244.140]^¼

T = 6.284 * (1.63*10^12)^¼

T = 6.284 * 1130

T = 7101 s

T = 118.35 mins

T = 1.9725 hrs

Answer

0.2067m or 0.2067m

Explanation;

Let lenght of spring= Lo= 11cm=0.110m

It is hang from a mass of

3.05-kg having a length of L1= 12.40cm= 0.124m

Force required to stretch the spring= Fkx

But weight of mass mg= kx then K= Mg/x

K= 3.05-kg× 9.8)/(0.124m-.110m)

K=2135N

But potential Energy U= 0.5Kx

X=√ 2U/k

√(2*10)/2135

X=0.0967m

The required new length= L2= L0 ±x

=

.110m ± 0.0967m

X= 0.2067m or 0.2067m hence the total lenghth

✴ Case - I

⟶ Force = F

⟶ Mass = m

⟶ Acceleration = 12m/s²

✴ Case - II

⟶ Force = 2F

⟶ Mass = 3m

➳ Acceleration in second case.

⇒ This question is completely based on the concept of newton's second law of motion.

⇒ As per this law, Force is defined as the product of mass and acceleration.

Mathematically, F = ma

[tex]\implies\sf\:\dfrac{F_1}{F_2}=\dfrac{m_1\times a_1}{m_2\times a_2}\\ \\ \implies\sf\:\dfrac{F}{2F}=\dfrac{m\times 12}{3m\times a_2}\\ \\ \implies\sf\:\dfrac{1}{2}=\dfrac{4}{a_2}\\ \\ \implies\sf\:a_2=4\times 2\\ \\ \implies\underline{\boxed{\bf{a_2=8\:ms^{-2}}}}[/tex]

New acceleration would be 12 m/s²

Given that;

Acceleration of object = 12 m/s²

New net force = 2f

New mass = 3m

Find:

New acceleration

Computation:

[tex]\frac{F1}{F2} = \frac{m1a1}{m2a2} \\\\\frac{f}{2f} = \frac{m(12)}{(3m)a2} \\\\\frac{1}{2} = \frac{4}{a2} \\\\a2 = 8 m/s^2[/tex]

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Answer:

14.29 turns.

Explanation:

From the question,

Ns/Np = Ip/Is........................ Equation 1

Where Ns = Secondary turn, Np = Primary turn, Is = current flowing in the secondary turn, Ip = current flowing in the primary turn.

Make Ns the subject of the equation

Ns = NpIp/Is.................... Equation 2

Given: Np = 100 turns, Ip = 2.0 A, Is = 14 A.

Substitute these values into equation 2

Ns = 100(2.0)/14

Ns = 14.29 turns.

Answer:

Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.

If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.

Answer:

Explanation:

We know that the equivalent resistance of resistors connected in parallel is expressed as

[tex]\frac{1}{Re} =\frac{1}{R1} +\frac{1}{R2}+\frac{1}{R3}\\\\\frac{1}{Re} =\frac{1}{R} +\frac{1}{2R}+\frac{1}{3R}\\[/tex]

the L.C.M is 6R

[tex]\frac{1}{Re} =\frac{6+3+2}{6R} = \frac{11}{6R} \\\\Re= \frac{6R}{11}[/tex]

Answer:

The new acceleration would be 9 m/s².

Explanation:

Acceleration of an object is 6 m/s²

Net force is equal to the product of mass and acceleration i.e.

F = ma

[tex]a=\dfrac{F}{m}\\\\\dfrac{F}{m}=6\ m/s^2[/tex]

If the net force was tripled and the mass were doubled, it means,

F' = 3F

m' = 2m

Let a' is new acceleration. So,

[tex]a'=\dfrac{F'}{m'}\\\\a'=\dfrac{(3F)}{(2m)}\\\\a'=\dfrac{3}{2}\times \dfrac{F}{m}\\\\a'=\dfrac{3}{2}\times 6\\\\a'=9\ m/s^2[/tex]

So, the new acceleration would be 9 m/s².

Answer:

a)  v = 2,9992 10⁸ m / s , b)  Eo = 375 V / m ,  B = 1.25 10⁻⁶ T,

c)     λ = 3,157 10⁻⁷ m,   f = 9.50 10¹⁴ Hz ,  T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

        E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

        v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

        k = 2π / λ

        λ = 2π / k

        λ = 2π / 1.99 10⁷

        λ = 3,157 10⁻⁷ m

        w = 2π f

        f = w / 2 π

        f = 5.97 10¹⁵ / 2π

        f = 9.50 10¹⁴ Hz

the wave speed is

        v = 3,157 10⁻⁷   9.50 10¹⁴

        v = 2,9992 10⁸ m / s

b) The electric field is

           Eo = 375 V / m

to find the magnetic field we use

           E / B = c

           B = E / c

            B = 375 / 2,9992 10⁸

            B = 1.25 10⁻⁶ T

c) The period is

           T = 1 / f

            T = 1 / 9.50 10¹⁴

            T = 1.05 10⁻¹⁵ s

the wavelength value is

          λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

Answer:

Explanation:

According to work energy theorem

change in kinetic energy of truck = work done against it

work done against it = force x displacement

= - 850 x 8 = 6800 J

change in kinetic energy of truck = - 6800 J .

energy will be reduced by 6800 J

Answer:-6800

Explanation:

Answer:

F = F₀ 0.2

Explanation:

For this exercise we apply Coulomb's law with the initial data

     F₀ = k q_A q_B / d²

indicate several changes

q_A ’= ½ q_A

q_B ’= 1/10 q_B

d ’= ½ d

let's substitute these new values ​​in the Coulomb equation

          F = k q_A ’q_B’ / d’²

          F = k ½ q_A 1/10 q_B / (1/2 d)²

          F = (k q_A q_B / d2) ½ 1/10 2²

          F = F₀ 0.2

Answer:

2.39 Ω

Explanation:

Given that

Number of winnings on the coil, = 250 turns

Radius if the copper wire, r(c) = 1.3/2 = 0.65 mm

Radius of single cylinder layer, R = 12 cm

Length of the cylinderical coil, L = 250 * 2π * 12 = 188.4 m

Resistivity of copper, ρ = 1.69*10^-8 Ωm

Area is πr(c)², which is

A = 3.142 * (0.65*10^-3)²

A = 3.142 * 4.225*10^-7

A = 1.33*10^-6 m²

The formula for resistance is given as

R = ρ.L/A, if we substitute, we have

R = (1.69*10^-8 * 188.4) / 1.33*10^-6

R = 3.18*10^-6 / 1.33*10^-6

R = 2.39 Ω.

Therefore, the resistance is 2.39 Ω

Answer:

Explanation:

Let the original resistance be R and voltage be V

Applying ohm's law

V / R = 15

V = 15 R

In second case

V / (R+8 ) = 12

V = 12 R + 96

15 R = 12 R + 96

3R = 96

R = 32 ohm .

Answer:

neither will happen

Explanation:

cause the water is already defreezed

Explanation:

work = force x Distance

w = 10 x 1.5 = 15Nm

The amount of work done by Kyla in lifting the bag is 15 J.

Work done on an object is defined as the cross product of the force applied on the object and the vertical displacement of the object.

Here,

Force applied by Kyla to pick up the bag, F = 10 N

Vertical displacement of the bag, s = 1.5 m

The work done by Kyla in lifting the bag,

W = F x s

W = 10 x 1.5

W = 15 J

Hence,

The amount of work done by Kyla in lifting the bag is 15 J.

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Answer:

The correct answer would be - dependent independent variable experiment.

Explanation:

Dr. Stein hypothesized that excess sugar causes hyperactivity, so sugar treatment /no sugar treatment would be independent variable. By giving some children sugar and others a sugar cookies he can manipulate the independent variable.

Similarly , the dependent variable is the result or outcome of independent variable, or what Dr. Stein hypothesize to be the result of excess sugar . In this sugar experiment, then, the dependent variable is the children's hyper activity level.

Thus, the correct answer would be - dependent independent variable experiment.

The best research method to use for the research of hyperactivity,  would be dependent-independent variable experiment.

The given problem is based on the effect of sugar on hyperactivity. Hyper activity refers to the increased movement, impulse actions and a shorter attention span.

Thus, we can conclude that the best research method to use,  would be - dependent-independent variable experiment.

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Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

Answer:

The frequency is    [tex]f = 0.221 \ Hz[/tex]

Explanation:

From the question we are told that  

     The  time taken for it to decay to half its original size is [tex]t = 3.40 \ ms = 3.40 *10^{-3} \ s[/tex]

Let the voltage of the capacitor when it is fully charged be  [tex]V_o[/tex]

Then the voltage of the capacitor at time t is  said to be  [tex]V = \frac{V_o}{2}[/tex]

   Now  this voltage can be  mathematical represented as

      [tex]V = V_o * e ^{-\frac{t}{RC} }[/tex]

Where  RC  is the time constant

   substituting values  

    [tex]\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }[/tex]

    [tex]0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }[/tex]

    [tex]- \frac{0.5}{RC} = ln (0.5)[/tex]

     [tex]-\frac{0.5}{RC} = -0.6931[/tex]

     [tex]RC = 0.721[/tex]

Generally the cross-over frequency for a low pass filter is mathematically represented as

          [tex]f = \frac{1}{2 \pi * RC }[/tex]

substituting values  

           [tex]f = \frac{1}{2* 3.142 * 0.72 }[/tex]

           [tex]f = 0.221 \ Hz[/tex]

Answer:

The speed of light will be c=3x10^8m/s

Explanation:

This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c

Answer:

The maximum displacement of the mass m₂ [tex]= \frac{2(m_1-m_2)g}{k}[/tex]

Explanation:

Kinetic Energy (K) = 1/2mv²

Potential Energy (P) = mgh

Law of Conservation of energy states that total energy of the system remains constant.

i.e; Total energy before collision = Total energy after collision

This implies that: the gravitational potential energy lost by m₁ must be equal to sum of gravitational energy gained by m₂ and the elastic potential energy stored in the spring.

[tex]m_1gd = m_2gd+\frac{1}{2}kd^2\\\\m_1g = m_2g+\frac{1}{2}kd\\\\d = \frac{2(m_1-m_2)g}{k}[/tex]

d = maximum displacement of the mass m₂