A block of mass \"m\" is compressed a distance \"s\" against a spring of force c

Question

A block of mass "m" is compressed a distance "s" against a spring of force constant "k" and released as show n at point A. The surface is frictionless until it reaches point "B". At point B the surface becomes rough. The coefficient of kinetic friction between the block and the surface is mu_1 and the length of the rough section is "2R". At point C the surface becomes frictionless again. The block then slides up the bowl shaped surface of radius "R" to a maximum angle of 90degree where it comes to rest at point D. Given [m, k, R, mu_k Calculater a.The speed of the block at point C. b.The speed of the block at point B. c.The initial compression of the spring. d.The work due to friction between A and D. e.The work due to gravity between A and D. f.The work due to the spring between A and D. g.The total work between A and D.

Explanation / Answer

(a)
Initial Spring Potential Energy = 1/2 * k * s^2
Using Energy Conservation,
At Point B,
Kinetic Energy = Initial Spring Potential Energy
1/2*m*vb^2 = 1/2 * k* s^2
vb = sqrt(k/m) * s ----------1

At point C
Initial K.E = Final K.E + Energy Lost in Friction
1/2*m*vb^2 = 1/2*m*vc^2 + m*g*uk*(2*R)
vb^2 = vc^2 + 2*g*uk*(2*R)
vc^2 = vb^2 -  4*g*uk*R ---------2

(c)
Now,
Using Energy Conservation we can say,
Initial Spring Potential Energy = Energy Lost in Friction + Gravitational Potential Energy
1/2 * k*s^2 = m*g*uk*(2*R) + m*g*R
s = sqrt((2*m*g*R (2*uk + 1))/k) ( Initial Compression)

(b)
Vb = sqrt(k/m) * sqrt((2*m*g*R (2*uk + 1))/k)
Vb = sqrt(2*g*R (2*uk + 1)) ( Speed at B)

(a)
Vc^2 = vb^2 -  4*m*g*uk*R
Vc^2 = 2*g*R (2*uk + 1) - 4*g*uk*R
Vc^2 = 2*g*R
Vc = sqrt(2*g*R) ( Speed at C)

Work due to Friction btw A & D, = 4*m*g*uk*R

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