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Naddika [18.5K]
3 years ago
8

How do the individual forces compare when the rider experiences a sensation being lighter than his normal weight ? What is the a

cceleration direction when this occurs ?
Physics
1 answer:
shusha [124]3 years ago
6 0

Answer:

When the elevator is accelerating downward, the person feels lighter due to the downward normal force being less than the person's weight.

Explanation:

A person riding in an elevator subjected to a series of unbalanced forces depending on the direction the elevator is travelling.

Two forces are acting on the person; the force of gravity and the upward normal force from the elevator.

When the elevator is going upwards with acceleration a, the person feels heavier than his normal weight, due to the upward normal force being greater than the person's weight. N = mg + ma

When the elevator is moving downwards with acceleration a, the person feels lighter due to the downward normal force being less than the person's weight. N = mg - ma

However, when the elevator is moving up or down at constant velocity ie. acceleration a = 0, the person experience a normal force equal to weight. N = mg

When the elevator is moving downwards with acceleration a = g, the person experiences weightlessness. N = (mg - mg) = 0

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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.15 s later. Ignor
pogonyaev

Answer:

a) v₀ = 9.2 m/s

b) y₀ = 7.9 m

Explanation:

The position of the balls is given by the equation:

y =- \frac{1}{2} gt^2 + v_0 t + y_0

where:

acceleration g = 9.8 m/s²

time t

initial velocity v₀

initial height y₀

a) lets divide (a) in two parts:

1.part: How long will it take the second ball to fall down?

v_0 = 0, y = 0\\0=- \frac{1}{2} gt^2 + y_0\\ t = \sqrt{\frac{2y_0}{g}}

2. part: At time t from part1 + 1.15s, the first ball should land on the ground.

y = 0, y_0 = 19.6, t = \sqrt{\frac{2y_0}{g}} + 1.15\\ 0 =- \frac{1}{2} gt^2 + v_0t + y_0

This leaves only one unknown: v₀

v_0 =\frac{1}{t}(\frac{1}{2} gt^2 - y_0)\\ v_0 = 9.2 \frac{m}{s}

b)again, lets divide in two parts

1.part: Where will ball1 be relative to ball2 in 1.15s:

t = 1.15s, v_0 = 8.6 m/s\\y= -\frac{1}{2} gt^2 + v_0t + y_0\\ \delta y = y - y_0 =v_0t -\frac{1}{2} gt^2

and how fast will it go:

v' = -gt + v_0

2.part: Now we can plug in to the equation for the position of the two balls. Let's start with the second ball first:

0 = -\frac{1}{2} gt^2 + y_0\\ y_0 = \frac{1}{2} gt^2

Now let's use this result in the equation for the first ball:

0 = - \frac{1}{2} gt^2 + v't + y_0 + \delta y = - \frac{1}{2} gt^2 + v't + \frac{1}{2} gt^2 + \delta y\\ 0 = v't + \delta y\\ t =- \frac{\delta y}{v'} \\ y_0 = \frac{1}{2} g(\frac{\delta y}{v'})^2\\ y_0 = 7.9m

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