Two objects of unequal mass (m_1 = 7 kg, m_2 = 15 kg) are hung vertically over a

Question

Two objects of unequal mass (m_1 = 7 kg, m_2 = 15 kg) are hung vertically over a frictionless pulley of negligible mass, as in shown in the figure. The magnitude of the acceleration of the two objects is 9.8 m/s^2 3.57 m/s^2 7.74 m/s^2 3.27 m/s^2 Based on the information in the previous problem, the Tension (T) in the cord is. 45.00 N 65.40 N 93.64 N 103.11 N A ball of mass 0.500 kg is attached to the end of a cord 1.0 m long. The ball is whirled in a horizontal circle. If the cord can withstand a maximum tension of 8.0 N, the maximum speed of the ball is then 3 m/s 2 m/s 6 m/s 4 m/s Based on the information in problem 11 the periodic tune of the ball when is rotated with the maximum speed is (pi/2) m/s (2pi/3) m/s (2pi/5) m/s (pi/3) m/s A sky diver of mass 60.0 kg jumps from an aircraft and reaches a terminal speed of 30.0 m/s. The drag force on the diver just when he reach this speed is 321.5 N 484.8 N 588.6 N 784.8 N

Explanation / Answer

9)

for m2

m2*g - T = m2*a

T = m2*g - m2*a .........(1)

for m1

T - m1*g = m1*a

m2*g - m2*a - m1*g = m1*a

a = (m2-m1)*g/(m1+m2)

a = (15-7)*9.81/(7+15)

a = 3.57 m/s^2

+++++++++++++

from 1

T = (15*9.81)-(15*3.57)

T = 93.6 N

++++++++++++++

centripetal force = T

Fc = T

m*v^2/L = T

0.5*v^2/1 = 8

v = 4 m/s

++++++++++++

time priod T = total distacne/v

T = 2*pi*L/v = 2*pi*1/4 = pi/2 s

+++++++++++++++++

under terminal motion

Fnet = 0

Fg - Fd = 0

Fd = Fg

Fd = mg

Fd = 60*9.81

Fd = 588.6 N

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