A block of mass M = 1 kg slides back and forth on a one-dimensional frictionless

Question

A block of mass M = 1 kg slides back and forth on a one-dimensional frictionless surface between two springs, of spring constant kleft = 30 N/m and kright = 48 N/m. When the block hits the left spring itcompresses a maximum distnace of dleft, max = 0.7 m.

1) What is the maximum compression of the right spring? _____ m

What is speed of the block when it is between the two springs? ________m/sec

Suppose instead that there is friction between the block and the surface and that the distance between the relaxed springs is d = 2.8 m. The block is initially placed against the left spring which is compressed by 0.7 m and then released. The first time the block hits the right spring, it is compressed by 0.43 m. Find the coefficient of kinetic friction between the block and the table. _____m/sec

PLEASE HELP

Explanation / Answer

here,

spring constant of spring's
kl = 30 N/m
kr = 48 N/m

mass of block, m = 1 kg
left spring compression, xl = 0.7

From Conservation of Energy,
PE of Spring on left = PE os spring on right
0.5 * kl *xl^2 = 0.5*kr*xr^2
0.5*30*0.7^2 = 0.5 *48*xr^2

solving for compression on right spring, xr
xr = sqrt( (30*0.7^2 )/48 )
xr = 0.553 m

Part B:
from conservation of Energy

kinectic erngy gaine dby block = pE stored in left or right spring
0.5 * m * v^2 = 0.5 * kl * xl^2

solving for velocity of block, v

v = sqrt(kl * xl^2 /m)
v = sqrt(30 *0.7^2/1)
v = 3.834 m/s

Ques.2
from work Energy theoram,
Work done = potential Energy lost by spring
Ff*d = 0.5 * kl*xl^2 ( Ff = Frictional Force)
2*u*mg*d = kl*xl^2

solving for coefficient of friction u

u = kl*xl^2 / (2*m*g)

u = 30*0.7^2/(2*1*9.8)

u = 0.75

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