Last time this was posted, the answer to part b was incorrect. Could someone els

Question

Last time this was posted, the answer to part b was incorrect. Could someone else give it a try? I've done the work for part a already.

A block of mass 1.3 kg is attached to a horizontal spring that has a force constant 900 N/m as shown in the figure below. The spring is compressed 2.0 cm and is then released from rest.

(a) A constant friction force of 3.6 N retards the block's motion from the moment it is released. How much is the spring compressed when the speed of the block is a maximum.

My answer: F = kx ---> x = F/k = 3.6N/900(N/m) = .004m = .4cm

(b) What is the maximum speed? (in cm/s)

Explanation / Answer

Solution:

Initial Energy
E = 1/2 k d²

At maximum speed
E' = 1/2 k x² + 1/2 m v²

Using Work energy theorem,

W = ?E

F (d - x) = 1/2 k d² - (1/2 k x² + 1/2 m v²)

v² = k d² / m - k x² / m - 2F (d - x) / m-----[1]

Now, Differentiate eq[1] and vmax if dv/dx = 0

2v dv/dx = - 2 k x / m + 2F / m = 0

F = k x

x = F / k

a) Now, using the above relation,

x = 3.6 / 900 = 4*10-3 m = 0.004m

x = 0.4 cm.

b)Not just substitute the x in eq[1] where k =900N/m , d = 0.02m ,m=1.3kg , F = 3.6N

v² = k d² / m - k x² / m - 2F (d - x) / m

= [900*0.022]/1.3 - [900*0.0042]/1.3 -[2*3.6*(0.02-0.004)]/1.3

= 0.17723 m/s

v = 0.421 m/s

v = 42.1 cm/s.

I hope you understood the problem, If yes rate me!! or else comment for a better solution.

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