Question

A block of mass m= 2.8 kg is attached to a spring of spring constant k= 500 N/m. the block is pulled to an initial position x = 5 cm to the right of the equilibrium point as shown below, and released from rest. Find the speed of the block as it passes through equilibrium if the coefficient of kinetic friction between the block and the surface μk = 0.35.

Answer 1

Answer:

Explanation:

The energy stored in a spring

= 1/2 k x²

where k is spring constant and x is extension in the spring.

= .5 x 500 x .05²

= .625 J

Work done by friction = energy dissipated

= - μmg x d , μ is coefficient of friction , m is mass , d is displacement

= - .35 x 2.8 x 9.8 x .05

= - .48 J

energy of the mass when it reaches equilibrium position

= .625 - .48

= .145 J

If v be its velocity at that time

1/2 m v ² = .145

.5 x 2.8 x v² = .145

v² = .10357

v = .32 m /s

32 cm /s