A 50 kg person is standing on a bathroom scale in an elevator. What force (in ne

Question

A 50 kg person is standing on a bathroom scale in an elevator. What force (in newtons) is indicated by the scale (a) If the elevator is not moving? (b) If the elevator is moving upward at a steady speed of 1.5 m/s? (c) If the elevator is accelerating upward at 2.2 m/s^2? (d) If the rising elevator decreases its speed from 0.50 m/s to 8.0 m/s in 3.0 seconds. Block m_1 in the figure below has a mass of 6 kg and m_2 has a mass of 1.4 kg. The coefficient of kinetic friction between m_2 and the horizontal surface is 0.55. The inclined plane is frictionless. (a) Find the tension in the cord and the acceleration of m_1. (b) If the coefficient of static friction is 0.65, what is the minimum of block m_2 that would prevent the system from moving.

Explanation / Answer

2. (a) When elevator isn't moving,

Force = mg = 50*9.8 = 490 N

(b) when the elevator is moving upward at a steady velocity, its acceleration is zero.

So, Force acting on the person = mg = 50*9.8 = 490 N

(c) If the elevator is accelerating upwards with an acceleration of 2.2 m/s^2

Force acting on the person = m(g-a) = 50(9.8-2.2) = 380 N

(d) Question is not accuarte. Let's take two cases,

i) If speed of elevator is going from 0.5 m/s to 8 m/s, it is increasing.

so, v = u+at

so, a = v-u/t = (8-0.5)/3= 2.5 m/s^2

then force, F = m(g-a) = 50(9.8-2.5)

= 365 N

ii) If the elevator speed is decreasing, it may be going from 8 m/s to 0.5 m/s

then acceleration = (0.5-8)/3 = -2.5 m/s^2

so, Force acting on the person = m(g-a) = 50*(9.8-(-2.5))

= 615 N

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