A lot of mass m2 = 28 kg hanging from a rope.
The rope runs over a frictionless pulley and attached to a box of mass m1 = 52 kg.
The box is on a plane with endpoints in A. Oblique Planet either by angle v = 180 in relation to the horizontal.
The friction coefficient between the box m1 plane is ì = 0.2.
The length of the plane from A to snaffle is great and always so large that the assumptions in the problem can be solved. The rope has always large enough length to keep the distance between the cage and pulley.
Assume that the rope is very smooth and slippery, and that both it and the pulley can be considered weightless (mass resolving) in this task.
Look away from time to tensioning of the rope,
assume that it is tight from the moment
It’s 4 question
1) The system is kept at rest and dropped to
motion arises or may arise.
Describe what happens when
system is released resolved, and calculate how
great power Ftau the rope is then.
2) After a time t = 3 seconds bust (cut) the rope.
What is the speed of the system then?
How far have the box and the weight shifted when this happens?
3) Will the box stop now as a result of the rope is cut? (Explain and show)
If so, for how long after the rope broke, and how
far, the fund total moved on before it stops?
4) Assume that there is sufficient free height of fall of the weight that it can fall freely throughout the time it takes to fund a halt.
How far has the lot fell in total - including the time it was tied to checkout
above?
Thanks for the help
To solve this problem, we will break it down into different steps:
Step 1: Determine the gravitational force acting on m1 and m2.
The gravitational force on an object is given by the equation F = m * g, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s^2). In this case, the mass of m1 is 52 kg, so the gravitational force on m1 is F1 = 52 kg * 9.8 m/s^2. The mass of m2 is 28 kg, so the gravitational force on m2 is F2 = 28 kg * 9.8 m/s^2.
Step 2: Calculate the tension in the rope.
Since the system is at rest initially, the tension in the rope is equal to the sum of the gravitational forces on m1 and m2. Therefore, the tension in the rope is T = F1 + F2.
Step 3: Determine the net force on the system.
The net force on the system can be calculated using the equation: Net Force = T - friction force. In this case, there is a friction force acting on m1 due to the incline. The friction force can be calculated using the equation: friction force = μ * F1, where μ is the friction coefficient (0.2) and F1 is the normal force on m1.
Step 4: Calculate the acceleration of the system.
Using Newton's second law (F = m * a), we can find the acceleration of the system: Net Force = (m1 + m2) * a.
Step 5: Determine the motion of the system when it is released.
If the net force on the system is greater than 0, then the system will begin to move. If the net force on the system is equal to 0, then the system will remain at rest. In this case, if the net force is greater than 0, the system will start moving.
Step 6: Calculate the power exerted by the rope.
Power is the rate at which work is done and can be calculated using the equation: Power = Force * Velocity. In this case, the power exerted by the rope can be calculated using the tension in the rope and the velocity of the system.
For the second part of the question, after the rope is cut, we can assume that there are no external forces acting on the system other than the weight of the masses. This means that the system will continue to move with a constant velocity. We can use the equations of kinematics to find the speed of the system and the distance traveled.
For the third part of the question, if the rope is cut, the box (m1) will experience a force due to friction with the inclined plane. This force will decelerate the box until it comes to a stop. The time it takes for the box to stop can be calculated using the equation: t = vf / a, where vf is the final velocity of the box and a is the deceleration.
For the fourth part of the question, if there is enough free height of fall for m1, we can calculate the distance it has fallen using the equation: d = 0.5 * g * t^2, where d is the distance, g is the acceleration due to gravity, and t is the time.
However, please note that in order to solve the problem accurately, we would need the values for the masses, angles, and coefficients of friction. Without these values, we can only provide a general explanation of how to approach the problem.
1) When the system is released, the weight of the hanging mass (m2) will cause the box (m1) to move downwards. The force of gravity acting on m2 will create a tension force in the rope. Since the rope is assumed to be very smooth and slippery, there is no friction between the rope and the pulley. Therefore, the tension force in the rope will be equal to the weight of m2.
The weight of m2 can be calculated using the formula: weight = mass * acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s^2.
weight of m2 = m2 * 9.8
= 28 kg * 9.8 m/s^2
= 274.4 N
So, the tension force in the rope (Ftau) is 274.4 N.
2) After the rope is cut, the box and the hanging mass will separate. The box will continue to move downwards due to the force of gravity, while the hanging mass will move upwards. The speed of the system can be calculated using the principle of conservation of mechanical energy.
The initial mechanical energy of the system is the sum of the potential energy and the kinetic energy:
Initial mechanical energy = potential energy (PE) + kinetic energy (KE)
Since the system is at rest initially, the initial kinetic energy is zero. The potential energy can be calculated using the formula: PE = mass * gravity * height.
For the box:
PE_box = m1 * 9.8 * distance_A_to_snaffle
For the hanging mass:
PE_hanging_mass = m2 * 9.8 * distance_A_to_snaffle
Since the potential energy is the same for both the box and the hanging mass, their sum is equal to the initial mechanical energy.
Initial mechanical energy = PE_box + PE_hanging_mass
When the rope is cut, the potential energy is converted into kinetic energy. Therefore, the final mechanical energy of the system is equal to the final kinetic energy.
Final mechanical energy = final kinetic energy
Since the kinetic energy is given by the formula: KE = (1/2) * mass * velocity^2, we can find the velocity of the system after the rope is cut using:
Final mechanical energy = (1/2) * (m1 + m2) * velocity^2
Solving this equation for velocity, we can find the speed of the system.
3) After the rope is cut, the box will not stop immediately. This is because the box is now subject to the force of kinetic friction between the box and the plane surface. The force of kinetic friction can be calculated using the formula: friction force = friction coefficient * normal force.
The normal force can be calculated by resolving the weight of the box into two components: one perpendicular to the plane surface (normal force) and one parallel to the plane surface (force of gravity).
normal force = m1 * gravity * cos(v)
friction force = μ * normal force
= μ * m1 * gravity * cos(v)
The friction force will act in the opposite direction to the motion of the box. Therefore, it will decelerate the box until it eventually comes to a stop.
The time after the rope is cut that it takes for the box to stop can be calculated using the equation: acceleration = (final velocity - initial velocity) / time
In this case, the initial velocity is the speed of the system calculated in step 2, and the final velocity is zero (since the box will come to a stop).
The distance the box moves after the rope is cut can be calculated using the formula: distance = initial velocity * time + (1/2) * acceleration * time^2.
4) If there is sufficient free height of fall for the hanging mass, it will fall freely under the action of gravity. The distance it falls can be calculated using the formula: distance = (1/2) * acceleration due to gravity * time^2.
Since the time it takes for the box to come to a stop after the rope is cut will be different from the time it takes for the hanging mass to fall freely, the total distance the hanging mass falls will be the sum of the distance it falls during free fall and the distance it has already moved before the rope was cut.