Question

AS shown in the figure, there is no friction between the horizontal surface and the lower block
(M = 3 kg) but friction coefficient between both the blocks is 0.2. Both the blocks move together with
al speed V towards the spring, compresses it and due to the force exerted by the spring, moves in
the reverse direction of the initial motion. What can be the maximum value of V (in cm/s) so that during
the motion, there is no slipping between the blocks (use g = 10 m/s2).
​

Answer 1

See Photo

Mark as Brainliest

Thanks

Answer 2

Given:

A system of two masses m and M with weight 1kg and 3kg respectively

A spring with a constant 400N/m

The friction coefficient between both the blocks = 0.2

To Find:

The maximum value of V (in cm/s) so that during the motion, there is no slipping between the blocks

Solution:

The maximum velocity at the condition of no slipping = 20cm/s

Let x be the displacement of the spring in meter and a be the acceleration of the block system.

F$_s$ = kx

= 400x

Using newton's equation,

F$_s$ = (m+M) a

or a = 400x/4

= 100x

For no slipping condition,

The frictional force ≥ The force on the upper block (ma)

or μmg ≥ ma

or 2 ≥ 100mx      - (1)

Applying the Work-Energy Theorem,

The net work done = ΔKE

Since the initial velocity of the system is 0 m/s,

ΔKE = 1/2 X (m+M) X v²

The work done due to the spring = 1/2 k x²

Equating,

(m+M) X v² = k x²

or x = $\sqrt{\frac{4v^2}{k} }$

Substituting the value of x in equation 1 at the limiting condition,

2= 100m$\sqrt{\frac{4v^2}{k} }$

or 1 = 50 $\sqrt{\frac{4v^2}{k} }$

Squaring both the sides, and substituting the value of k

400 = 2500 X 4v²

or 1 = 25v²

or v = 1 /√25

= 1/ 5 ms⁻¹

= 20 cm/s      (1m = 100cm)