A 2 kg block is connected into a spring attached to a wall with spring constant

Question

A 2 kg block is connected into a spring attached to a wall with spring constant of 12 N/m. The spring is compressed a distance of 1m, and the block is released in the horizontal direction. The static friction coeficient between the block and the surface is 0.6 and the kinetic friction coeficient is 0.3

(a) How far does the block travel before coming to rest?

(b) What's the time it takes to travel from the point of release to the point where it comes to rest?

Please can you show the concepts and/or formulas used to solve it. I have to understand it because I have an exam on this subject.

Thanks!

Explanation / Answer

initially x = 1m

K = 12N/m therefore Fsp = Kx = 12N

force of static friction = Us*N = Fs

N = normal reaction force = M*g = 20N

therefore Fs = 12N since Us=0.6 for static friction

now if we compress the spring by a small dX further then the force Kx becomes more than the maximum static friction force hence it begins to move .

now kinetic friction begins to act

Fk =Uk*N = 6N since Uk = 0.3

therefore now a constant Fk will act

therefore net force on the block = Kx - Fk = Kx - 6 = M*a where a = accn of the block

thus it will be a S.H.M with time period = squareroot(m/k)*2pi

T = 2.56 seconds

T = time taken by the block to return back to the position from where it started motion therefore time taken to come to rest will be T/2 = 1.28 seconds

to calculate the distance covered by block before it comes to rest we can use a very smart concept which involves finding the distance covered at maximum velocity from zero veocity because at maximum velocity the accn = 0 therefore

Kx - 6 = 0 therefore x = 0.5m

the distance taken to come to rest will be double this distance = 1m

thus it comes to rest at 1m distance

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