An elevator is being pulled up from the ground floor to the third floor by a cable. The cable is exerting 4500 newtons of force

Answer:

a) The ball's velocity just prior to hitting the ground measures -6.3 m/s

b) The ball's velocity right after bouncing off the ground registers at 3.1 m/s

c) The average acceleration's magnitude is 470 m/s², and its direction is upward, forming a 90º angle with the ground.

Explanation:

To begin, let’s assess the time it takes for the ball to reach the floor:

The equation outlining the ball's position is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at given time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration triggered by gravity

We establish the ground as the reference origin.

a) Since the ball is released rather than thrown, the initial velocity v0 is 0. The direction of acceleration is downward, directed towards the origin; thus, “g” is treated as negative. When the ball contacts the ground, its position will be 0. Therefore:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²

-2.0 m = -4.9 m/s² * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The motion equation for a falling body is:

v = v0 + g * t      where "v" denotes the velocity

Since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) The pebble's speed reaches 0 during its maximum height. To find the time taken for the pebble to achieve that height, we can use the velocity equation and then substitute that time in the position equation to derive the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

Replacing t in the position equation, knowing the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration can be determined by:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

The direction of the acceleration is upward, perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

Answer:

Insufficient details provided; please clarify further.

a)

i) 120 s

ii) 1.57 m/s

b)

i) Refer to the attached diagram

ii) Up

c)

d) Greater than

Explanation:

The problem does not provide full details: consult the attachments for the complete text.

a)

The revolution period of the book equals the total duration needed for the book to make one full revolution.

By examining the graph, we can approximate the revolution period by calculating the time difference between two successive points of the book's motion that share the same shape.

We could use the time difference between two adjacent crests to estimate the period. The first crest is observed at t = 90 s, and the following crest appears at t = 210 s.

This results in the revolution period being

T = 210 - 90 = 120 s

ii)

The tangential speed of the book is computed as the ratio of the distance traveled over one revolution (i.e., the circumference of the wheel) to the revolution period.

Mathematically:

where

R represents the wheel radius

T = 120 s indicates the period

Based on the graph, the book reaches a maximum at x = +30 m and a minimum at x = -30 m, giving the diameter of the wheel as

d = +30 - (-30) = 60 m

This means the radius calculates to

R = d/2 = 30 m

So, the final speed is

b)

i) Please consult the attached free-body diagram for the book when at its lowest point.

Two forces act on the book at the lowest position:

- The weight of the book, represented as

where m denotes the book's mass and g stands for gravitational acceleration. This force functions downward.

- The normal force the bench exerts on the book is represented by N. This force acts upward.

ii)

While at its lowest position, the book maintains a horizontal motion at constant speed.

Nevertheless, the book is undergoing acceleration. Acceleration is defined as the rate of velocity change, which is vectorial, having both speed and direction. While the speed remains unchanged, the direction changes (upward), indicating the book has upward net acceleration.

According to Newton's second law, the net vertical force acting on the book corresponds with the vertical acceleration:

where F = net force, m = mass, a = acceleration. Thus, if a is non-zero, the upward net force must exist in line with the direction of the acceleration.

c)

As discussed in part b), there are two forces influencing the book at the lowest point:

- The weight, , directed downward

- The normal force from the bench, N, directed upward

Given that the book is in uniform circular motion, the net force must match the centripetal force , leading us to the equation:

where

represents the speed of the book

R stands for the radius of the circular path.

We derive an expression for the normal force:

d)

As per the discussions in parts c) and d):

- The normal force acting on the book at its lowest point becomes

- The weight (gravitational force) of the book is

Upon comparing these two equations, we conclude:

Thus, it is evident that the normal force exerted by the bench exceeds the weight of the book.