A massless spring constant \"k\" is fixed on the left side of a level track. A b

Question

A massless spring constant "k" is fixed on the left side of a level track. A block of mass "m" is pressed against the spring and compresses it a distance of "d". The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius "R". The entire track and the loop-the-loop are frictionless, expect for the section of track between points A and B. Given that the coefficient of Kinetic Friction between the block and the track along AB is "mk" and that length of AB is "L".
A) Calculate the speed "Va" of the mass as it passes point A in terms of "d".
B) Calculate the speed "Vb" of the mass as it passes point B in terms of "Va"
C) Calculate the speed "Vtop" of the mass at the top of the loop in terms of "Vb"
D) What is the normal force on the mass at the top of the loop (in terms of "Vb")?

Explanation / Answer

(a) Apply conservation of energy when the spring is in compressed postion and just after released.                          (1/2)kd2 = (1/2)mva2                                  va = ( kd2 / m )1/2 (b)Since there is no other forces other than the frictional force, then acceleration of the block is                 a = f / m                    = -g     From equation of motion                   vb2 - va2 = 2aL                   vb2 - va2 = 2(-g)L                    vb2 = va2 - 2gL                    vb = ( va2 - 2gL )1/2 (c) Apply conservation of energy at bottom and at top of the loop,                 (1/2) mvb2 = mg(2R) + (1/2) mvtop2                            vb2 = 4gR + vtop2                           vtop2 =  vb2  - 4gR                             vtop = ( vb2  - 4gR )1/2 (d) Normal force on the block at the top of the loop is                  N + mg = mvtop2 / R                           N = mvtop2 / R - mg    (b)Since there is no other forces other than the frictional force, then acceleration of the block is                 a = f / m                    = -g     From equation of motion                   vb2 - va2 = 2aL                   vb2 - va2 = 2(-g)L                    vb2 = va2 - 2gL                    vb = ( va2 - 2gL )1/2 (c) Apply conservation of energy at bottom and at top of the loop,                 (1/2) mvb2 = mg(2R) + (1/2) mvtop2                            vb2 = 4gR + vtop2                           vtop2 =  vb2  - 4gR                             vtop = ( vb2  - 4gR )1/2 (d) Normal force on the block at the top of the loop is                  N + mg = mvtop2 / R                           N = mvtop2 / R - mg                      vb2 - va2 = 2(-g)L                    vb2 = va2 - 2gL                    vb = ( va2 - 2gL )1/2 (c) Apply conservation of energy at bottom and at top of the loop,                 (1/2) mvb2 = mg(2R) + (1/2) mvtop2                            vb2 = 4gR + vtop2                           vtop2 =  vb2  - 4gR                             vtop = ( vb2  - 4gR )1/2 (d) Normal force on the block at the top of the loop is                  N + mg = mvtop2 / R                           N = mvtop2 / R - mg    (c) Apply conservation of energy at bottom and at top of the loop,                 (1/2) mvb2 = mg(2R) + (1/2) mvtop2                            vb2 = 4gR + vtop2                           vtop2 =  vb2  - 4gR                             vtop = ( vb2  - 4gR )1/2 (d) Normal force on the block at the top of the loop is                  N + mg = mvtop2 / R                           N = mvtop2 / R - mg    (d) Normal force on the block at the top of the loop is                  N + mg = mvtop2 / R                           N = mvtop2 / R - mg   

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