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:0.9 m/s

Explanation:

Use work energy theorem,

Work done = ∆K.E

Work done= work done by spring +

Work done by friction

=-(-1/2kx^2)+(-N×S)

Answer 2

Solution:-When the spring is expanded upto .005m from it's initial position

then,force applied by spring on the block=$kx$

here,k=500N/m,,x=0.05m

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

Now,when the block is released from that position,and started to move toward its initial position

in that case

force towards it's motion=25N

force against it's motion=frictional force=u×N

=.35×28=9.8

resultant force=25-9.8=15.7N

acceleration in block=force/mass=15.7/2.8=5.6m/s^2

now,

when the block is released

distance to travel (to it's initial position)=0.05

acceleration=5.6m/sec^2

initial velocity=0

using third law of uniformly accelerated motion,we can say

V^2=0^2+2×5.6×.05

V^2=.56

v=7.4m/s

hence, when the block will be at it's initial position then,it's velocity will he 7.4m/s

{hope it helps you}