Question

Fill in the Blank :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 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 . . . . . . .

Answer 1

Answer:

The value of N is 4

Explanation:

Let the problem the mass of the the block is m which is attached with a spring which has a force constant let,k and the coefficient friction μ = 0.1The displacement of the block from its initial position to final due to impulse , x = 0.06 m

It is given that the initial velocity of the block is V = N/10

Now, loss in Kinetic energy = gain in potential energy + work done

1/2 mV^2 = 1/2kx^2 +  μgx

V = √k/mx^2 +  μgx

  = √2/0.18 x (0.06)^2 +0.1 x 10 x 0.06

  = 4/`0

Therefore putting the value of V

V = N/10

N = 4