Dimas Pyrros (four-time Olympic medalist) raises a 215 kg barbell in two stages. In the
first stage he raises it to his shoulders, a height of 1.60 m, by exerting a maximum force of
3000 N. In the second stage he raises the barbell above his head at a constant speed to
a maximum height of 2.10 m. He then holds it above his head for 2.0 s before dropping it
to the mat.
(a) What is the magnitude of the force of gravity on the barbell?
(b) How much work did Pyrros do on the barbell to raise it to a height of 1.60 m?
(c) What is the magnitude of the force which the weightlifter must apply to raise the
barbell the last 0.60 m?
(d) How much work did the weightlifter do on the barbell to raise it the last 0.60 m?
(e) Draw a FBD of the barbell while it is held aloft for 2.0 s
(f) How much work did the weightlifter do on the barbell during those 2.0 s?
Hello,
I need to know the answer to the part (c) to (f).
thanks for the help
c, constant speed means force up = weight = m g in Newtons
d, weight in Newtons * 0.60 meters = work done in Joules
e, force up = weight down, no acceleration
f, force times distance = ZERO because no distance.
(a) The magnitude of the force of gravity on the barbell can be calculated using the formula:
Force of gravity = mass x gravitational acceleration
First, we need to find the mass of the barbell. To do this, we can use the equation:
Weight = mass x gravitational acceleration
Given that Pyrros is able to raise the barbell, we can assume that the weight of the barbell is equal to the maximum force he exerts (3000 N). So,
Weight = 3000 N
Next, we need to convert the weight to mass. The gravitational acceleration on Earth is approximately 9.8 m/s^2.
Weight = mass x 9.8 m/s^2
Solving for mass:
mass = weight / 9.8 m/s^2 = 3000 N / 9.8 m/s^2 = 306.12 kg
Therefore, the magnitude of the force of gravity on the barbell is equal to its weight, which is 3000 N.
(b) The work done by Pyrros to raise the barbell to a height of 1.60 m can be calculated using the equation:
Work = force x distance
Given that the maximum force exerted by Pyrros is 3000 N and the height raised is 1.60 m, we can calculate:
Work = 3000 N x 1.60 m = 4800 J
Therefore, Pyrros did 4800 Joules of work on the barbell to raise it to a height of 1.60 m.
(c) To find the magnitude of the force which the weightlifter must apply to raise the barbell the last 0.60 m, we can use the equation:
Force = mass x gravitational acceleration
The mass of the barbell remains the same at 306.12 kg, and the gravitational acceleration is 9.8 m/s^2.
Force = 306.12 kg x 9.8 m/s^2 = 2996.78 N
Therefore, the magnitude of the force which the weightlifter must apply is approximately 2996.78 N.
(d) The work done by the weightlifter to raise the barbell the last 0.60 m can be calculated using the equation:
Work = force x distance
Given that the magnitude of the force is 2996.78 N and the distance raised is 0.60 m, we can calculate:
Work = 2996.78 N x 0.60 m = 1798.07 J
Therefore, the weightlifter did approximately 1798.07 Joules of work on the barbell to raise it the last 0.60 m.
(e) The free-body diagram (FBD) of the barbell while it is held aloft for 2.0 s would show the following forces acting on the barbell:
- The force of gravity (3000 N) acting downward
- The force exerted by the weightlifter (equal in magnitude and opposite in direction to the force of gravity) acting upward
(f) The work done by the weightlifter on the barbell during those 2.0 s can be calculated using the equation:
Work = force x distance
Given that the force exerted by the weightlifter and the distance (height) raised above the head are constant, we can calculate:
Work = force x distance = 3000 N x 2.10 m = 6300 J
Therefore, the weightlifter did 6300 Joules of work on the barbell during those 2.0 seconds.
To find the answers to the given questions, we can use principles of mechanics and apply relevant formulas. Let's go through each of the questions step by step:
(a) The magnitude of the force of gravity on the barbell can be found using the equation:
Force of gravity = mass * acceleration due to gravity
The mass of the barbell is not given explicitly, but we can use Newton's second law of motion to find it. The force exerted in the first stage and the resulting acceleration can be related as:
Force = mass * acceleration
Rearranging the equation, we get:
mass = Force / acceleration
From the given information, the force in the first stage is 3000 N. The acceleration due to gravity is approximately 9.8 m/s².
Therefore, the mass of the barbell is:
mass = 3000 N / 9.8 m/s²
Calculate this to find the mass. Once we have the mass, we can calculate the force of gravity using the first equation.
(b) The work done by Pyrros on the barbell to raise it to a height of 1.60 m can be calculated using the work-energy principle:
Work = force * distance * cos(theta)
In this case, the force is the force exerted in the first stage (3000 N), the distance is the height (1.60 m), and theta is the angle between the direction of the force and the displacement (which is 0 degrees since both are in the same direction).
The magnitude of the force of gravity on the barbell is acting downward, so the angle is 180 degrees. Plug in the values and calculate the work.
(c) To calculate the magnitude of the force the weightlifter must apply to raise the barbell the last 0.60 m, we can use the principle of mechanical equilibrium. When the barbell is raised at a constant speed, the force applied by the weightlifter is equal and opposite to the force of gravity acting on the barbell.
The magnitude of the force the weightlifter must apply is equal to the magnitude of the force of gravity on the barbell, which can be calculated in part (a).
(d) The work done by the weightlifter to raise the barbell the last 0.60 m can be calculated using the work-energy principle, similar to part (b). The force to be used is the magnitude of the force in part (c), the distance is 0.60 m, and theta is also 180 degrees since force and displacement are in opposite directions.
(e) To draw a free body diagram (FBD) of the barbell while it is held aloft for 2.0 s, we need to consider the forces acting on it. The main forces are the force of gravity (acting downward) and the force applied by the weightlifter (acting upward).
(f) The work done by the weightlifter on the barbell during those 2.0 s can be calculated using the work-energy principle. The force to be used is the magnitude of the force in part (c), the distance is the vertical displacement of the barbell (2.10 m), and theta is 180 degrees since force and displacement are in opposite directions.
Calculate the work using the given values to find the answer.
By following these steps, you can find the answers to each of the questions.