A 79.0 kg man stands on a spring scale in an elevator. Starting from rest, the e

Question

A 79.0 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.11 m/s in 1.30 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 2.00 s and comes to rest. (What does the spring scale register before the elevator starts to move? (b)What does it register during the first 1.30 s? (What does it register while the elevator is traveling at constant speed? (d)What does it register during the time it is slowing down? In the Atwood machine shown below, m_1 = 2.00 kg and m_2 = 7.50 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at = 2.80 m/s downward. (How far will m_1 descend below its initial level? (Find the velocity of m_1 after 1.80 s.

Explanation / Answer

part 1)

a ) at rest w = mg

w = 79 * 9.8 = 774.2 N

part b )

for first 1.3s

a = 1.11/1.3

w = m(g+a)

w = 841.65 N

part c )

at constant velocity a = 0

w = mg

w = 774.2 N

part d )

coming downward

a = 1.11/2

w = m(g-a)

w = 730.355 N

part 2 )

for atwwod machine

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

a = 5.67368 m/s^2

vf^2 = vi^2 + 2a*s

vf = 0

s = (0 -vi^2)/2a

s = 0.69 m

part b )

v = u + at

v = -2.80 + 5.67368*t

t = 1.8s

v = 7.412624 m/s ( upward direction )

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