Physics - Unit V Challenge Problem

No matter how bad your day is going, here's proof that things could be A LOT worse.
This is a bricklayer's accident report that was printed in the newsletter of the English
equivalent of the Workers' Compensation Board. So here, thanks to John Sedgwick, is this
Bricklayer's report.
Dear Sir;
I am writing in response to your request for additional information in Block #3 of the
accident reporting form. I put "Poor Planning" as the cause of my accident. You asked for a
fuller explanation and I trust the following details will be sufficient. I am a bricklayer by
trade. On the day of the accident, I was working alone on the roof of a new six-story
building. When I completed my work, I found I had some bricks left over which when
weighed later were found to weigh 240 lbs. Rather than carry the bricks down by hand, I
decided to lower them in a barrel by using a pulley which was attached to the side of the
building at the sixth floor. Securing the rope at ground level, I went up to the roof, swung
the barrel out and loaded the bricks into it. Then I went down and untied the rope, holding it
tightly to insure a slow descent of the 240 lbs of bricks. You will note on the accident
reporting form that my weight is 135 lbs.
Due to my surprise at being jerked off the ground so suddenly, I lost my presence of mind
and forgot to let go of the rope. Needless to say, I proceeded at a rapid rate up the side of
the building.(Q1) In the vicinity of the third floor, I met the barrel which was now
proceeding downward at an equally impressive speed. (Q2) This explains the fractured
skull, minor abrasions and the broken collarbone, as listed in Section 3, accident reporting
form.
Slowed only slightly, I continued my rapid ascent, not stopping until the fingers of my right
hand were two knuckles deep into the pulley which I mentioned in Paragraph 2 of this
correspondence. Fortunately by this time I had regained my presence of mind and was able
to hold tightly to the rope, in spite of the excruciating pain I was now beginning to
experience.
At approximately the same time, however, the barrel of bricks hit the ground-and the bottom
fell out of the barrel. Now devoid of the weight of the bricks, the barrel weighed
approximately 50 lbs. I refer you again to my weight. As you might imagine, I began a rapid
descent down the side of the building. (Q3) In the vicinity of the third floor, I met the
barrel coming up. (Q4) This accounts for the two fractured ankles, broken tooth and severe
lacerations of my legs and lower body.
Here my luck began to change slightly. The encounter with the barrel seemed to slow me
enough to lessen my injuries when I fell into the pile of bricks and fortunately only three
vertebrae were cracked. I am sorry to report, however, as I lay there on the pile of bricks, in
pain, unable to move and watching the empty barrel six stories above me, I again lost my
composure and presence of mind and let go of the rope. (Q5, 6, 7)In order to do these problems, you need to know that 1 lb ≈ 4.5 N. Convert the weights
involved to newtons. Assume that the 6-floor building is 20 meters high.
1. Draw a force diagram detailing the forces acting on the man and on the barrel of bricks
(290 lbs). Determine the acceleration of the system..
2. How fast was the barrel traveling when it struck the man 1/2 way up the building? What
was the barrel's velocity relative to the man?
3. Draw a new force diagram for the barrel and man after the bricks have fallen out.
Determine the new acceleration of the system.
4. How fast was the barrel traveling when it struck the man a 2nd time? What was its
velocity relative to the man?
5. Once the man let go of the rope, assume that the barrel was essentially in free fall.
(Assume negligible air resistance). How long did it take the barrel to reach the ground?
6. How fast was it going when it hit the man?
7. How fast would it have been going if it had been full of bricks (290 lbs, not 50 lbs)?
Explain your answer.

fwb=1305

fwp= 608N (force weight of person)

Fwb= 1305N (force wight of barrel)
FN= 1305-608=697N
F=ma
W=mg

This is a start sorry I cant help more.

1. To draw the force diagram, we need to consider the forces acting on the man and the barrel. For the man, there are two main forces: his weight (Wman) and the tension force in the rope (Tman). The weight of the man can be calculated by multiplying his mass (mman) by the acceleration due to gravity (g). Assuming the weight of the man is 135 lbs, we can convert it to Newtons by multiplying by the conversion factor 4.5 N/lb:

Wman = mman * g
Wman = (135 lbs) * (4.5 N/lb)
Wman = 607.5 N

The tension force acting on the man can be calculated using Newton's second law (F = ma):

Tman - Wman = mman * a

For the barrel, there are also two main forces: the weight of the barrel (Wbarrel) and the tension force in the rope (Tbarrel). The weight of the barrel can be calculated using the weight of the bricks (240 lbs), which we also need to convert to Newtons:

Wbarrel = mbarrel * g
Wbarrel = (240 lbs + 50 lbs) * (4.5 N/lb)
Wbarrel = 1350 N

The tension force acting on the barrel is the same as the tension force acting on the man.

Now, let's calculate the acceleration of the system. Since the rope is fixed at the sixth floor, the acceleration of the system will be the same for both the man and the barrel. Therefore:

Tman - Wman = mman * a
Tbarrel - Wbarrel = mbarrel * a

Substituting the known values:

Tman - Wman = (135 lbs) * a
Tbarrel - Wbarrel = (290 lbs) * a

Converting the weights to Newtons:

Tman - Wman = 607.5 N * a
Tbarrel - Wbarrel = 1305 N * a

Now, since the system is in equilibrium (the man and the barrel are not accelerating vertically), we can set these two equations equal to each other and solve for the acceleration (a):

607.5 N * a = 1305 N * a

Dividing both sides by a:

607.5 N = 1305 N

Since this equation cannot be true, the system is not in equilibrium. Therefore, the acceleration (a) is zero.

2. Since the acceleration of the system is zero, the barrel is traveling at a constant velocity when it strikes the man halfway up the building. The barrel's velocity relative to the man is also zero.

3. After the bricks have fallen out, the weight of the barrel reduces to 50 lbs. We can calculate the new weight of the barrel in Newtons:

Wbarrel = mbarrel * g
Wbarrel = 50 lbs * (4.5 N/lb)
Wbarrel = 225 N

The force diagram for the barrel and man after the bricks have fallen out will only include the tension force in the rope (Tbarrel/man) and the new weight of the barrel (Wbarrel).

4. Since the acceleration of the system is still zero after the bricks have fallen out, the barrel's velocity remains the same when it strikes the man for the second time. The barrel's velocity relative to the man is zero.

5. Once the man let go of the rope, the barrel is essentially in free fall, assuming negligible air resistance. The time it takes for the barrel to reach the ground can be calculated using the equation of motion:

s = ut + (1/2)at^2

Here, the initial velocity (u) is zero, the acceleration (a) is due to gravity (g), and the distance (s) is the height of the building (20 meters).

20m = (1/2) * g * t^2
40m = 9.8m/s^2 * t^2
t^2 = 40m / 9.8m/s^2
t^2 = 4.08s^2

Taking the square root of both sides:

t ≈ 2.02s

Therefore, it takes approximately 2.02 seconds for the barrel to reach the ground.

6. Since we know the height of the building is 20 meters and it takes the barrel approximately 2.02 seconds to reach the ground, we can calculate its velocity when it hits the man using the equation of motion:

v = u + at

Here, the initial velocity (u) is zero, the acceleration (a) is due to gravity (g), and the time (t) is 2.02 seconds.

v = 0 + 9.8m/s^2 * 2.02s
v ≈ 19.8 m/s

Therefore, the barrel is traveling at a velocity of approximately 19.8 m/s when it hits the man.

7. If the barrel had been full of bricks (weighing 290 lbs), we can calculate its new weight in Newtons:

Wbarrel = 290 lbs * (4.5 N/lb)
Wbarrel = 1305 N

Using the same equation of motion as before, we can calculate the velocity of the barrel when it hits the man:

v = u + at

Here, the initial velocity (u) is zero, the acceleration (a) is due to gravity (g), and the time (t) is 2.02 seconds.

v = 0 + 9.8m/s^2 * 2.02s
v ≈ 19.8 m/s

Therefore, regardless of whether the barrel is full or empty, its velocity is approximately 19.8 m/s when it hits the man. This is because the acceleration due to gravity is constant for all objects in free fall, regardless of their mass.

To solve these physics problems, we need to analyze the forces acting on the system and use Newton's laws of motion. Let's go step by step to find the answers:

1. Draw a force diagram detailing the forces acting on the man and the barrel of bricks (290 lbs). Determine the acceleration of the system.

a. Forces acting on the man:
- Weight (mg) acting downward
- Tension in the rope (upward force)

b. Forces acting on the barrel:
- Weight (mg) downward
- Tension in the rope (upward force)

To determine the acceleration, we need to equate the net force to the mass times acceleration (F = ma). The net force is the difference between the tension in the rope and the weight:

For the man: Tension - Weight(man) = ma(man)
For the barrel: Tension - Weight(barrel) = ma(barrel)

First, convert the weights from pounds to newtons:
Weight(man) = 135 lbs * 4.5 N/lb
Weight(barrel) = 290 lbs * 4.5 N/lb

Next, substitute the values into the equations above and solve for tension:

Tension - (Weight(man)) = m(man) * a (man)
Tension - (Weight(barrel)) = m(barrel) * a (barrel)

The values of m(man) and m(barrel) will be given in the problem statement. Use the equations to find the acceleration.

2. To find how fast the barrel was traveling when it struck the man halfway up the building and its velocity relative to the man, we need to use the equation of motion:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Given that the initial velocity is zero (because the barrel was initially at rest) and the displacement is half of the building's height (i.e., 10 meters), we can solve for the final velocity.

v^2 = 0 + 2a * 10

Using the known value of acceleration from the previous step, solve for the final velocity.

3. Draw a new force diagram for the barrel and man after the bricks have fallen out. Determine the new acceleration of the system.

After the bricks have fallen out, the only force acting on the barrel is its weight. So the forces acting on the barrel are:
- Weight (mg) downward.

The force diagram for the man remains the same as in the first step

To determine the new acceleration, we follow the same steps as in the first step but use the weight of the barrel (50 lbs * 4.5 N/lb) as a downward force acting on the barrel.

4. To find how fast the barrel was traveling when it struck the man a second time and its velocity relative to the man, we can use the same equation of motion as in step 2.

Plug in the new acceleration (obtained in the previous step) and the displacement (remaining height of the building) to solve for the final velocity.

5. Once the man let go of the rope, assume that the barrel was essentially in free fall (negligible air resistance). To determine how long it took for the barrel to reach the ground, we can use the kinematic equation:

s = ut + 1/2 at^2

where s is the distance covered, u is the initial velocity (in this case, zero), a is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.

Since the barrel falls freely, the distance covered is the height of the building (20 meters). Solve the equation for t to find the time taken.

6. To find how fast the barrel was going when it hit the man after the man let go of the rope, we can use the equation of motion from step 2 and substitute the value of time (obtained in step 5).

7. To calculate how fast the barrel would have been going if it had been full of bricks (290 lbs, rather than 50 lbs), we can use the principle of conservation of mechanical energy.

The initial potential energy when the barrel was full of bricks is equal to the final kinetic energy when it hits the man. The potential energy can be calculated as the product of the weight (290 lbs * 4.5 N/lb) and the height of the building (20 meters). Then, equate this potential energy to the kinetic energy, using the mass of the barrel (290 lbs) and the velocity obtained in step 6. Solve for the new velocity.

Remember to convert the weights from pounds to newtons, as given in the problem statement.

By following these steps, you should be able to solve the physics problems related to the given scenario.