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Question: In the situation given, the reading of the spring balance is (\(g = 10m/{s^2}\)). ![](https://www....

In the situation given, the reading of the spring balance is (g=10m/s2g = 10m/{s^2}).

A). 30N30N
B). 33N33N
C). 39N39N
D). 40N40N

Explanation

Solution

In order to find the solution of the question, we need to draw a free body diagram of the above image including all the forces exerted on the body. Then we will find the tension in the spring balance to find the reading.

Complete step-by-step solution:

The free body diagram of the body is given as below:
The frictional force exerted by 2kg2kg of body is
f1=μN{f_1} = \mu {\rm N}
f1=μmg=0.5×2×10 =10N{f_1} = \mu mg = 0.5 \times 2 \times 10 \\\ = 10N
The frictional force exerted by 3kg3kg body is
f2=μN f2=μmg =0.5×3×10 =15N{f_2} = \mu {\rm N} \\\ \Rightarrow f_2^{} = \mu mg \\\ = 0.5 \times 3 \times 10 \\\ = 15N
Then we need to find the acceleration find the tension between the spring:
Fnet=ma 60N+10N15N15N=(3+2)a 5a=40 a=8m/s2{F_{net}} = ma \\\ \Rightarrow 60N + 10N - 15N - 15N = (3 + 2)a \\\ \Rightarrow 5a = 40 \\\ \Rightarrow a = 8m/{s^2}
Now, we will find the tension in the wire of spring of the 2kg2kgbody:
T1+f115N=2a T1+1015=2×8 T1=21N{T_1} + {f_1} - 15N = 2a \\\ \Rightarrow {T_1} + 10 - 15 = 2 \times 8 \\\ \Rightarrow {T_1} = 21N
Now, we will find the tension in the wire of spring of the3kg3kgbody:
60NT2f2=3a 60NT215N=24 T2=19N60N - {T_2} - {f_2} = 3a \\\ \Rightarrow 60N - {T_2} - 15N = 24 \\\ \Rightarrow {T_2} = 19N
Now the reading of spring balance will be:
=T1+T2 =21N+19N =40N= {T_1} + {T_2} \\\ = 21N + 19N \\\ = 40N
The reading in the spring balance in the above figure is 40N40N.

Note: We need to keep some points in our mind while solving these problems:

The free body diagram is the most important thing in these kinds of questions so that we get to know the magnitude and direction of all the forces exerted.
We should have proper knowledge of laws of motion so that we can apply the accordingly.
We should remember all the required formulae.