Question

In the given diagram, reading of spring balance will be( g= 10m/s^2)​

Answer 1

Given:

A spring balance with 2 masses have been provided.

To find:

Reading of spring balance ?

Calculation:

• Always remember that SPRING BALANCE READS THE TENSIONS OF THE STRING.

So, let the tension be T :

For 6 kg mass :

$6g - T = 6a$

$\implies \: 60 - T = 6a$

For 3 kg mass:

$T - 3g = 3a$

$\implies \: T - 30 = 3a$

Now, add the equations :

$\implies \: 60 - 30 = 6a + 3a$

$\implies \: 9a = 30$

$\implies \: a = \dfrac{10}{3} \: m {s}^{ - 2}$

Now, tension in string is :

$T - 30 = 3a$

$\implies \: T = 30 + 3 \times \dfrac{10}{3}$

$\implies \: T = 30 + 10$

$\implies \: T = 40 \: N$

So, spring balance reading will be 40 N (option 2)

Answer 2

Answer:

Tension in spring is 40N

OR

Reading of spring is 40N

Explanation:

Given :

block 1 = 6 kg

block 2 = 3 kg

g = 10$m/s^{2}$

To find : Tension in spring

Solution : Given figure has two blocks of 6kg and 3kg. As the weight of block 1 is greater than block two, g of block 1 is positive and g of block 2 is negative.

Let T be the tension and a be acceleration for both blocks.

For block 1,

6g - T = 6a

∴ T = 6g - 6a .......................... (equation 1)

For block 2,

T - 3g = 3a

∴ T = 3a + 3g .......................... (equation 2)

Now, equate equation 1 and equation 2

we get,

6g - 6a = 3g + 3a

∴ 6g - 3g = 6a + 3a

∴ 3g = 9a

∴ a = $\frac{g}{3}$    $m/s^{2}$

∴ a = $\frac{10}{3}$ $m/s^{2}$

Now, Tension in the spring is

T - 3g = 3a

i.e. T = 3a +3g

T = 3* $\frac{10}{3}$ + 3 * 10

∴T = 10+ 30

∴ T = 40N

∴ Tension in the spring is 40 N.

#SPJ3