Question

If F1 is 20N ,F2 is 10N and m1 is 4kg ,m2 is 2kg .the tension in the spring connected between blocks is

Answer 1

Answer:

First take both blocks together

F=MA

20-10 = 6 x a

a= 10/6 =5/3ms-2

Now for 2 kg

T-10 =2x 5/3

T=10+10/3= 40/3 N

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Answer 2

Question :-

If F1 is 20N ,F2 is 10N and M1 is 4kg ,M2 is 2kg .The tension in the spring connected between blocks is ?

Given :-

• Force 1 (F1) → 20 N
• Force 2 (F2) → 10 N
• Mass 1 (M1) → 4 Kg
• Mass 2 (M2) → 2 Kg

Formula :-

• The only formula required is F = Ma .

Solution :-

• Firstly , take / bring both F1 , F2 in the two sides like forces and Both the Masses M1 , M2 in between the forces .

$\sf10N \leftarrow4 + 2 \rightarrow20N$

Now, applying the formula,

F = Ma

• Here to find F we need to substract F2 from F1
• And, to find M we need to add M1 and M2 .
• And as because we don't know the value of Acceleration (a) so, we will keep it like that only .

$\implies \sf20 - 10 = 4 + 2 \times a$

$\implies \sf20 - 10 = 6 \times a$

$\implies \sf 10 = 6 \times a$

$\implies \sf a= \cancel\dfrac{10}{6}$

$\implies \sf a= \dfrac{5}{3} m/s ^{2}$

Now , read the question once more carefully it is asked for 2 kg .

Now, we know the value of a so,

$10 \sf \leftarrow2 a\rightarrow \: t$

$\sf \: t - 10 = 2 \times \dfrac{5}{3}$

$\sf \: t =\dfrac{10}{3} + 10$

$\sf \: t =\dfrac{10}{3} +3\times 10$

$\sf \: t =\dfrac{10}{3} + 30$

$\sf \: t =\dfrac{40}{3} N$

$\sf \: \therefore The \:value\: of \:t\: =\dfrac{40}{3} N$