The acceleration of gravity is 9.8 m/s2 . To emphasize the fact that lessons lea

Question

The acceleration of gravity is 9.8 m/s2 .
To emphasize the fact that lessons learned
in 303K carry over into real life, this prob-
lem is based on an accident report from
the Worker’s Compensation Board in Bryant,
Texas.
Dear Sir/Madam:
I am writing in response to your request
for a fuller explanation on the cause of my
accident. I trust the following details will
be sufficient. On the day of the accident, I
was working alone laying bricks on the roof
of a new building. On completing my work,
I had some bricks left over which were found
to have a mass of 101 kg. Rather than carry
the bricks down by hand, I decided to lower
them in a barrel with a mass of 20.3 kg by
using a pulley which was attached to the side
of the building at the roof. 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 bricks. You will note that my mass is
73.8 kg. To my surprise, I was jerked off
the ground suddenly and forgot to let go of
the rope. Needless to say, I proceeded at a
rapid rate up towards the roof of the building.
Somewhere up the building, I met the barrel
which was now proceeding downward at an
equally impressive speed. This did not slow
me down and I continued my rapid ascent,
hitting the pulley on the top floor. I estimate
that all this happened in the matter of 2.81 s.
At the same time, however, the barrel of bricks
hit the ground and the bottom fell out of the
barrel. Devoid of the bricks, the barrel now
had a mass of 20.3 kg. I refer you again to
my mass. As you might imagine, I began
a rapid descent down the side of the building
and met the barrel coming up. My luck finally
began to change slightly. The barrel slowed
me enough to lessen my injuries when I fell
into the pile of bricks. 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 several stories above me, I again lost
my composure and let go of the rope. These
events account for the fractured skull, broken
collarbone, fractured ankles, broken tooth,
severe lacerations and cracked vertebra.
Sincerely,
I am A. Klutz

a)What was the initial upward acceleration
experienced by the worker? Answer in units
of m/s2.
b)What is the height of the building? Answer
in units of m.
c)How many seconds did it take (after starting
his descent) for the worker to hit the ground?
Answer in units of s.

Explanation / Answer

Well, you certainly get some hilarious word problems in this physics class! However, it is actually not all that difficult, so don't get overwhelmed. First, skim the problem and find the important information:

m(bricks) = 101 kg
m(barrel) = 20.3 kg
m(person) = 73.8 kg

t = 2.81 s

Next, draw free body diagrams for both the man and the barrel, and label the forces on both. Remember to add the tension in the rope, which will be upward, as well as the force of gravity.

a) From Newton's 2nd Law, you know that:

a = Fnet / m(net)

Fnet = F(gravity on barrel) - F(gravity on man) = g*(m(barrel) + m(bricks)) - g*m(man)

= 9.81*(101+20.3) - 9.81*78.3 = 421.83 N

m(net) = m(bricks) + m(barrel) + m(man) = 101 + 20.3 + 73.8 = 199.6 kg

Plug this in, and you get:

a = F(net)/m(net) = 421.83/199.6 = 2.39 m/s^2 going up

b) To find the height, we need to use a kinematics equation:

y = y0 + v0*t + 0.5*a*t^2

Since we are starting at the ground from rest, y0 = 0 and v0 = 0, so this becomes:

y = 0.5*a*t^2

They gave us t, and we just found a, so now we just plug it all in to find the vertical distance the man went, which was the height of the building:

y = 0.5*2.39*(2.81)^2 = 9.44 m

c) Since the mass of the barrel is now only 20.3 kg, you need to redo your acceleration calculations from part a:

Fnet = 9.81*20.3 - 9.81*78.3 = -568.98 N

m(net) = 20.3 + 78.3 = 98.6 kg

a = -568.98/98.6 = -5.77 m/s^2

Now we use the kinematics equation again. This time y(initial) = 9.44 m, y(final) = 0, a = -5.77 m/s^2, and we want to find t.

0 = 9.44 - 0.5*5.77*t^2

Solving for t^2, we get:

t^2 = 9.44/(0.5*5.77) = 3.272

Taking the square root of this, our final answer is 1.81 seconds

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