Question

Three blocks S, R and P connected by string
are arranged as shown in the figure. The
masses of S, R and P are 6 kg, 6 kg and 12 kg
respectively. The tension in the string
connecting R and P is (ignore friction)
I
S
21​

Answer 1

Given that,

Mass of block of S = 6 kg

Mass of block of R = 6 kg

Mass of block of P = 12 kg

Suppose a force of 24 N is applied to the string connected to P

We need to calculate the acceleration

Using newton's second law

$F=ma$

Put the value into the formula

$24=(6+6+12)a$

$a=\dfrac{24}{24}$

$a=1\ m/s^2$

We need to calculate the tension in the string connecting R and P

Using balance equation

$24-T_{1}=12a$

$-T_{1}=-24+12$

$T_{1}=12\ N$

Hence, The tension in the string connecting R and P is 12 N.

Answer 2

The value of acceleration is 1 m/s^2b and tension is 12 N.

Explanation:

• Mass of "S" block = 6 Kg
• Mass of "R" block = 6 Kg
• Mass of "P" bloc = 12 Kg

Solution:

Now using the second law of newton to find acceleration.

The equivalent mass is = [ 6 + 6 + 12 ] = 24 Kg

F = ma

24 N = 24 Kg x a

a = F / m

a = 24 / 24 = 1 m/s^2

Now for tension.

F - T = 12 a

24 - T = 12 a

24 - T = 12( 1)

24 - 12 = T

T = 12 N

Hence the tension is 12 N.