Question

a block of mass m is attached to a light spring of force cosntant k. the block is placed on a rough incline for which u=3/4.initially block os at rest nd spring is unstretched. min value of M to moce the mass m up the incline is? (M is attached by a string to the other end of the sprin and it hanging vertically off the incline

Answer 1

M = m (sin θ + (3/4) cos θ)

Answer 2

We start by considering an infinitesimal displacement of the block upward along the plane. When the block is about to move, all forces are in (almost) static balance with friction at its maximum.

Let the block move a distance x (along the incline) so that a spring extension of x gives a spring force of kx. Because the string attached to the mass M is inextensible and assumed to be always taut, the hanging mass M will move downward by x.

For the block of mass m on the incline the forces along the plane are:

• Up the plane: Spring force = kx
• Down the plane: Component of weight = m g sinθ and friction = µ m g cosθ (with µ = 3/4)

At the threshold of motion the net force on m is zero:

kx = m g sinθ + (3⁄4)m g cosθ  (1)

For the hanging mass M (assuming equilibrium at the threshold) the spring force equals its weight:

kx = M g  (2)

Equate (1) and (2):

M g = m g sinθ + (3⁄4)m g cosθ

Cancel g:

M = m (sinθ + (3/4) cosθ)

Thus, the minimum value of M required to initiate upward motion is

M = m (sinθ + (3/4) cosθ).

Minimal Explanation of the Solution

1. Set up force balance on m at the moment of impending motion:
   Spring force kx must overcome mg sinθ + µ m g cosθ.

2. Relate spring force to hanging mass:
   Since the spring is light and the string taut, kx = M g.

3. Equate and solve:
   M g = m g (sinθ + µ cosθ) → M = m (sinθ + (3/4) cosθ).