Question

Two blocks A and B connected by an ideal spring of constant 100 N/m and moving on a smooth horizontal surface under influence of a force of 36 N. What is extension in spring when acceleration of 2 kg block is twice that of 5 kg block?​

Answer 1

Answer: here’s the answer, all the best with studying

Explanation:

Answer 2

Answer:

the extension in spring when the acceleration of 2kg and 5 kg is 0.1m

Explanation: we know that the force is known by the mass× acceleration(A)

given that: mass of A =5kg

mass of B =2 kg

spring constant (k)=100 N/m

and the horizontal force: 36N

F=Kx (this is the spring force ).....................(1)

and the force applied on the block (b)  F=M(b) × A .................(2)

So, we can say that Kx=M(b)×A ................... from equations (1 and 2)

and the force applied to the block (a) is ,

F(net)=35-Kx

M(a)×A = 35-Kx..........................(3)

M(a)× A = 35- M(b)×A

M(a)×A + M(b) × A = 35

M(a) + M(B) × A =35

A= 35/7    (mass of block (a) and mass of block (b) is 5 and 2 )

A = 5m/s^2

now putting acceleration in equation  1

Kx = M(b) × A

100x = 2×5

x = 0.1 m

so, the extension in the spring is 0.1m

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