Riding in an elevator, a man notices that he feel heavier or lighter based when the elevator speeds up or slows down. He stands on a bathroom scale in the elevator to take some measurements. Assume that his mass is 85 kg.

a. What is this man’s weight while at rest?

b. What does the scale read when the elevator accelerates upward at 2.35 m/s2?

c. What is the acceleration (magnitude AND direction) of the elevator if the scale reads 700 N?

d. What acceleration (magnitude AND direction) is required for the man to feel “weightless”? Prove it by showing all work.

Part A:

weight at rest = 85 kg

Part B:
9.8 m/s² - 2.35 m/s² = 7.45 m/s²
W = (7.45 m/s²)(85 kg) = 633.25 N

Part C:
a = F/m = (700 N)/(85 kg) = 8.24 m/s² (downward)

Part D:
to feel "weightless," acceleration must be 9.8 m/s²
9.8 m/s² - 9.8 m/s² = 0 m/s²
W = (0 m/s²)(85 kg) = 0 N

Correction:

In Part C, it should say upward, not downward.

Herp: Weight is in newtons, not kg.

a) weight at rest: mg=85*9.8 N
b) weight=m(g+a)=85(9.8+2.35)=1033N
c) Net weight= m(g+a), a positive direction is up
745=m(g+a)
745/85 = g+a
745/85-g=a =-1.04m/s, direction negative means it is going downward
d)
to be weightless, the elevator must be falling at 9.8 m/s
Net weight=m(g+a)
a=-g

a. The man's weight while at rest can be calculated using the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. Assuming the acceleration due to gravity is 9.8 m/s^2, the weight of the man can be calculated as follows:

W = (85 kg) * (9.8 m/s^2)
W = 833 N

Therefore, the man's weight while at rest is 833 N.

b. When the elevator accelerates upward at 2.35 m/s^2, we need to find the new weight of the man. Since the elevator is accelerating upwards, the net force acting on the man increases. The formula to calculate the weight in this case is:

W = (mass * acceleration due to gravity) + (mass * acceleration of the elevator)

W = (85 kg * 9.8 m/s^2) + (85 kg * 2.35 m/s^2)
W = 833 N + 199.75 N
W = 1032.75 N

Therefore, the scale would read 1032.75 N when the elevator accelerates upward at 2.35 m/s^2.

c. To find the acceleration of the elevator, we can rearrange the formula from part b as follows:

acceleration of the elevator = (W - (mass * acceleration due to gravity)) / mass

acceleration of the elevator = (700 N - (85 kg * 9.8 m/s^2)) / 85 kg
acceleration of the elevator = (700 N - 833 N) / 85 kg
acceleration of the elevator = -0.5647 m/s^2

The negative sign indicates that the acceleration of the elevator is downward.

Therefore, the acceleration of the elevator, both in magnitude and direction, is 0.5647 m/s^2 downward.

d. To feel weightless, the man must experience zero net force. This means that the upward force acting on the man (provided by the scale) must equal the force of gravity acting downward.

Let's denote the acceleration required for the man to feel weightless as "a". The equation to calculate the scale reading is:

W = (mass * acceleration due to gravity) + (mass * acceleration of the elevator)

Since the man is weightless, the scale reading should be zero. Therefore, we have:

0 = (85 kg * 9.8 m/s^2) + (85 kg * a)

Simplifying the equation, we get:

-840.5 m/s^2 = 85 kg * a

Dividing both sides by 85 kg, we find:

a = -9.9 m/s^2

The negative sign indicates that the acceleration required for the man to feel weightless is downward.

Therefore, the acceleration required for the man to feel weightless, both in magnitude and direction, is 9.9 m/s^2 downward.