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

Concept:-Law of conservation of energy will be used to solve this question:-

Solution:-

When the string is expanded to upto 0.05m,then

force applied by spring on block=kx=500×.05=25N

When,the block starts to move ,then,force which oppose it's movement=frictional force=.35×2.8=.98=1(approx)

resultant force on block during it's movement=25-1=24N

acceleration=resultant force/mass=24/2.8=2.5

now,when the block is released

initial velocity=0

acceleration=2.5

distance=.05m

use the third Las of uniformly accelerated motion

${v}^{2} = 0 + 2 \times 2.5 \times .05 \\ {v}^{2} = .25 \\ v = .5$

Hence,the velocity with which the block will pass through the equilibrium position will be 0.5m/s

{hope it helps you}