Answer in Mechanics | Relativity for happy45 #108239

Explanation

• It is the pulled spring that supplies the energy for the block to move forward initially. Thereafter that energy is employed in gaining some kinetic energy and working against the frictional force generated due to the relative motion between the floor & the block.
• After returning to the O, block possesses some momentum which is distributed among itself & the lump of clay at the next step.
• The equilibrium velocity of the clay-block system need to be found using the “conservation of the linear momentum” concept.
• The affect of the frictional force in this case could be neglected due to the negligible time & the direction of the frictional force.
• For the new movement, dynamic frictional force is applied changed due to increment of the mass.
• With all of these, the new motion takes place in exhausting the kinetic energy attained, against the frictional force & in to the spring.

Notations

• V : velocity gained by the block.
• V₁: velocity of the clay-block system
• f : frictional force on the block
• f₁ : frictional force on the clay-block system
• m & m₁ : mass of block & clay-block system respectively

Calculations

1). Considering energy conservation, A to O.

“begin{aligned}nsmallfrac{1}{2}kx^2& =small f x+frac{1}{2} m v^2\nsmallfrac{1}{2}k x^2& =small (mu_km g ) x+frac{1}{2} mv^2\nsmall 0.5sdot 2500sdot 0.2^2&=small(0.3sdot 24)sdot 0.2,+,0.5sdot 2.4sdot v^2\nsmall bold v&=small bold{6.36 ,ms^{-2}}nend{aligned}”

2). Considering the linear momentum along the moving direction,

“qquadnbegin{aligned}nsmall mv&=small m_1v_1\nsmall2.4sdot 6.36&= small (2.4+0.25)sdot v_1\nsmall bold {v_1}&= small bold {5.76, ms^{-2}}nend{aligned}”

Now, the changed frictional force,

“qquadnbegin{aligned}nsmall f_1 &= small mu_1 m_1g\n&=small bold {7.95,N} nend{aligned}”

Then considering the energy conservation from O to B,

“qquadnbegin{aligned}nsmall frac{1}{2}m_1v_1^2&= small f_1d, + , frac{1}{2}kd^2\nsmall 43.96&= small 7.95d,+,1250d^2\nsmall 1250d^2, +, 7.95d-43.96&=0\nsmall bold d&=small bold { 0.184, m}nend{aligned}”

3). Now (consider system at rest as in sketch no. iv) the force expelled on clay-block system by the compressed spring is,

“qquadnbegin{aligned}nsmall F_{spring}&=small kd\n&= small bold {460N} nend{aligned}”

Now the idea is, if the maximum static frictional force at B is greater than the spring force, the clay-block would move or the negativity.

Therefore, the frictional force,

“qquadnbegin{aligned}nsmall f_{s.maximum}&= small mu_sm_1g\n&=small 0.4sdot (2.4,+,0.25)sdot 10\n&=small bold{10.6N ,(<F_{spring})}nend{aligned}”

Therefore, clay-bloc would move after compressing at B.