A block of mass 0.18 kg is attached to a spring of force-constant 2 N / m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m / s is V = N / 10 . Then N is

Question

A block of mass 0.18 kg is attached to a spring of force-constant 2 N / m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m / s is V = N / 10 . Then N is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Applying work energy theorem -(1/2) kx 2-μ mg x =-(1/2) m V 2 ⇒ V =(4/10) N =4

Applying work energy theorem −21​kx2−μmgx=−21​mV2 ⇒V=104​ N=4

Applying work energy theorem
$- \frac{1}{2} k x^{2} - μ m gx = - \frac{1}{2} m V^{2}$
$\Rightarrow V = \frac{4}{10}$
$N = 4$