Γ145 A Body Si Of Mass M=1kg Is Left At The Vertex A Of A Vertical Quadrant Of Radius R=12m. The Body, Reaching The Base Of The Quadrant, Continues To Move By Sliding On The Horizontal Surface Of A Table Of Height H=5m And Length 7.2m. The Body Si,
Introduction
In this article, we will delve into a complex motion scenario involving a body of mass 1kg, which is left at the vertex of a vertical quadrant of radius 12m. The body reaches the base of the quadrant and continues to move by sliding on the horizontal surface of a table of height 5m and length 7.2m. We will apply the principles of conservation of energy and momentum to analyze the motion of the body and determine its final velocity.
The Initial Motion
The body is initially at rest at the vertex A of the vertical quadrant. As it begins to roll down the quadrant, its potential energy is converted into kinetic energy. The potential energy of the body at the vertex A is given by:
U = mgh
where m is the mass of the body, g is the acceleration due to gravity, and h is the height of the vertex above the base of the quadrant.
U = 1kg x 9.8m/s^2 x 12m = 117.6J
As the body rolls down the quadrant, its potential energy is converted into kinetic energy, which is given by:
K = (1/2)mv^2
where v is the velocity of the body at the base of the quadrant.
K = (1/2) x 1kg x v^2
At the base of the quadrant, the body has a velocity v, which can be determined by equating the potential energy at the vertex A to the kinetic energy at the base of the quadrant:
U = K
117.6J = (1/2) x 1kg x v^2
v^2 = 235.2
v = 15.32m/s
The Motion on the Horizontal Surface
The body continues to move by sliding on the horizontal surface of the table. As it moves, its kinetic energy remains constant, but its potential energy changes due to the change in height. The potential energy of the body at a height h above the base of the table is given by:
U = mgh
where m is the mass of the body, g is the acceleration due to gravity, and h is the height of the body above the base of the table.
U = 1kg x 9.8m/s^2 x h
As the body moves along the table, its potential energy changes due to the change in height. The body starts at a height of 5m above the base of the table and moves to a height of 0m at the end of the table.
The Work Done by Friction
As the body moves along the table, it experiences a force of friction, which opposes its motion. The work done by friction is given by:
W = F x d
where F is the force of friction and d is the distance over which the force is applied.
W = F x 7.2m
The force of friction is given by:
F = μ x N
where μ is the coefficient of friction and N is the normal force.
F = 0.5 x 1kg x 9.8m/s^2 = 4.9N
W = 4.9N x 7.2m = 35.28J
The Change in Kinetic Energy
As the body moves along the table, its kinetic energy changes due to the work done by friction. The change in kinetic energy is given by:
ΔK = W
ΔK = 35.28J
The Final Velocity
The final velocity of the body can be determined by equating the initial kinetic energy to the final kinetic energy:
K = (1/2)mv^2
K = (1/2) x 1kg x v^2
v^2 = 235.2
v = 15.32m/s
However, we must also consider the work done by friction, which reduces the kinetic energy of the body. The final kinetic energy is given by:
K = K - ΔK
K = (1/2) x 1kg x v^2 - 35.28J
v^2 = 200.92
v = 14.17m/s
Conclusion
In this article, we have analyzed the motion of a body of mass 1kg, which is left at the vertex of a vertical quadrant of radius 12m. The body reaches the base of the quadrant and continues to move by sliding on the horizontal surface of a table of height 5m and length 7.2m. We have applied the principles of conservation of energy and momentum to determine the final velocity of the body.
References
- [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of physics. John Wiley & Sons.
- [2] Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers. Cengage Learning.
Note
Q: What is the initial potential energy of the body at the vertex A of the vertical quadrant?
A: The initial potential energy of the body at the vertex A of the vertical quadrant is given by:
U = mgh
where m is the mass of the body, g is the acceleration due to gravity, and h is the height of the vertex above the base of the quadrant.
U = 1kg x 9.8m/s^2 x 12m = 117.6J
Q: What is the velocity of the body at the base of the quadrant?
A: The velocity of the body at the base of the quadrant can be determined by equating the potential energy at the vertex A to the kinetic energy at the base of the quadrant:
U = K
117.6J = (1/2) x 1kg x v^2
v^2 = 235.2
v = 15.32m/s
Q: What is the potential energy of the body at a height h above the base of the table?
A: The potential energy of the body at a height h above the base of the table is given by:
U = mgh
where m is the mass of the body, g is the acceleration due to gravity, and h is the height of the body above the base of the table.
U = 1kg x 9.8m/s^2 x h
Q: What is the work done by friction as the body moves along the table?
A: The work done by friction as the body moves along the table is given by:
W = F x d
where F is the force of friction and d is the distance over which the force is applied.
W = F x 7.2m
The force of friction is given by:
F = μ x N
where μ is the coefficient of friction and N is the normal force.
F = 0.5 x 1kg x 9.8m/s^2 = 4.9N
W = 4.9N x 7.2m = 35.28J
Q: What is the change in kinetic energy of the body as it moves along the table?
A: The change in kinetic energy of the body as it moves along the table is given by:
ΔK = W
ΔK = 35.28J
Q: What is the final velocity of the body after it moves along the table?
A: The final velocity of the body after it moves along the table can be determined by equating the initial kinetic energy to the final kinetic energy:
K = (1/2)mv^2
K = (1/2) x 1kg x v^2
v^2 = 235.2
However, we must also consider the work done by friction, which reduces the kinetic energy of the body. The final kinetic energy is given by:
K = K - ΔK
K = (1/2) x 1kg x v^2 - 35.28J
v^2 = 200.92
v = 14.17m/s
Q: What are the assumptions made in this analysis?
A: The calculations presented in this article are based on the assumption that the body is a perfect sphere and that the surface of the table is smooth and frictionless. In reality, the body may experience some frictional forces, which would affect its motion. Additionally, the calculations assume that the body starts at rest at the vertex A of the vertical quadrant. In reality, the body may have some initial velocity, which would affect its motion.
Q: What are the limitations of this analysis?
A: The analysis presented in this article is a simplified model of a complex motion scenario. In reality, the motion of the body may be affected by various factors, such as air resistance, surface roughness, and initial velocity. Therefore, the results of this analysis should be interpreted with caution and should not be used to make predictions about real-world motion scenarios.
Q: What are the applications of this analysis?
A: The analysis presented in this article has applications in various fields, such as physics, engineering, and robotics. It can be used to model and analyze complex motion scenarios, such as the motion of a robot arm or the motion of a projectile. It can also be used to design and optimize systems that involve complex motion, such as conveyor belts or amusement park rides.