1) When a mass M=0.50 kg is suspended from a vertical spring, the spring\'s leng

Question

1) When a mass M=0.50 kg is suspended from a vertical spring, the spring's length increases by 7.50 cm. The mass M is now pulled downward an additional 2.00 cm; after the mass is released, the spring-mass oscillacts in SHM. Ignore all friiction losses.

Find the spring constant k

2) A 0.125 kg meter is supported by two strings- the first string is at A (the 15.0 cm mark) and the second string is at B (the 60.0 cm mark).

c)  Prove that the sum of the torques about the 50.0 cm mark (C) is zero

d) Prove that the sum of torques about the 0 cm mark (O) is zero.

Explanation / Answer

1)

k*x = M*g

k*0.075 = 0.5*9.81

so.. spring constatnt k = 65.4 N/m

2)

tenision in A = T_A = 0.2725 N upward

tenision in B = T_B = 0.95375 N upward

c) taking torque of all the forces about C ..

torque = T_A*(50-15)/100 - T_B*(60-50)/100 = 0.2725*0.35 - 0.95375*0.10 = 0

hence proved

d) torque about 0 ...

torque = T_A*(15)/100 + T_B*(60)/100 - W*50/100 = 0.2725*0.15 + 0.95375*0.60 - 0.125*9.81*50/100 = 0

hence proved

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