the t h op radius R 13 cm The blocks released from res a point with 5R above the

Question

the t h op radius R 13 cm The blocks released from res a point with 5R above the bottom of the loop. What are the magnitudes of (a) the horizontal component and b the vertical component of the net force acting on the block at point At what height h should the block be released from rest so that it is on the verge of losing contact with the track at the top of the loop? (On the verge of losing contact means that the normal force on the block from the track has just then become zero) o 0.047 kg can slide along the friction ess loop the In the figure, a small block of mass m TBP (a) Number o.24 (b) Number 0.39 (c) NumberT 0.78 Units TTJ Units TTJ UnitsTT J

Explanation / Answer

Using law of conservation of energy

Total energy at P = Total Energy at Q

(m*g*h) = (m*g*R)+(0.5*m*v^2)

m cancels

(9.8*5*0.13) = (9.8*0.13)+(0.5*v^2)

v = 3.2 m/s

a) Horizontal component of force is Fx = m*v^2/R = 0.047*3.2^2/0.13 = 3.7 N

b) vertical component of force is Fy = m*g = 0.047*9.8 = 0.461 N

c) at the top ,for not loosing contact

m*v^2/R = m*g

v = sqrt(R*g)

then

Total energy at P = Total energy at the top of the loop

m*g*h = (0.5*m*v^2)+(m*g*2R)

m cancels

(9.8*h) = (0.5*0.13*9.8)+(9.8*2*0.13)

h = 0.325 m = 32.5 cm

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