Problem 8 - A Complete Resource Book in Physics for JEE Main

Question

A block of mass \(m\) is attached to a massless spring of spring constant \(K\). This system is accelerated upward with acceleration \(a\). The elongation in spring will be (A) \(\frac{m g}{K}\) (B) \(\frac{m(g-a)}{K}\) (C) \(\frac{m(g+a)}{K}\) (D) \(\frac{m a}{K}\)

Step-by-Step Solution

Answer

The elongation in the spring will be \(\frac{m(g+a)}{K}\) (Option C).

Step 1: (Step 1: Define the forces on the block)

Firstly, let's define the forces acting on the block: 1. Gravitational force: F_g = m*g, acting downward 2. Spring force: F_s = K*x, acting upward (x is the elongation in the spring) 3. Inertial force: F_i = m*a, acting downward (as the system is accelerated upward)

Step 2: (Step 2: Equate the forces for equilibrium)

As the system is in equilibrium (no net force), the sum of all forces acting on the block is equal to zero: F_s - F_g - F_i = 0

Step 3: (Step 3: Substitute the forces)

Now, let's substitute the forces from step 1 into the equation from step 2: K*x - m*g - m*a = 0

Step 4: (Step 4: Solve for x)

Solve the equation for x, the elongation in the spring: K*x = m*g + m*a x = \(\frac{m(g+a)}{K}\) The elongation in the spring will be \(\frac{m(g+a)}{K}\) (Option C).