Conservation of Energy Equations -

[jpond89]

1.) A bead slides without friction around a loop-the-loop as shown in the figure below. The bead is released from a height h = 4.00R.
(a) What is the bead's speed (V) at point A? Answer in terms of R and g, the acceleration of gravity.

(b) How large is the normal force on the bead if its mass is 5.10 g?


W_net=K_F - K_I

Sorry I have no idea what is going on in this class. I am trying really hard to understand but my teacher is very old, he is wise, but it is really hard to understand him and he writes really small so I can't understand it. He assigned everyone seats and I was forced to sit in the back. Please help me.


2.) A particle of mass m = 3.90 kg is released from point A and slides on the frictionless track shown in the figure below. (ha = 6.70 m.)
(a) Determine the particle's speed at points B and C.
(b) Determine the net work done by the gravitational force in moving the particle from A to C.



3.) A light, rigid rod is 76.6 cm long. Its top end is pivoted on a low-friction horizontal axle. The rod hangs straight down at rest with a small massive ball attached to its bottom end. You strike the ball, suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle?



4.) The coefficient of friction between the 3.00 kg block and surface in the figure below is 0.300. The system starts from rest. What is the speed of the 5.00 kg ball when it has fallen 1.90 m?



5.)A 5.20 kg block is set into motion up an inclined plane with an initial speed of v0 = 7.70 m/s. The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal.
(a) For this motion, determine the change in the block's kinetic energy.
(b) For this motion, determine the change in potential energy of the block-Earth system.
(c) Determine the frictional force exerted on the block (assumed to be constant).
(d) What is the coefficient of kinetic friction?



6. A loaded ore car has a mass of 895 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 29.0° above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 13.0 s and then continues at constant speed.
(a) What power must the winch motor provide when the car is moving at constant speed?
(b) What maximum power must the winch motor provide?
(c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1350 m?



7. A 10.0 kg block is released from point A in the figure below. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2200 N/m, and compresses the spring to 0.350 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C.



8. A 20.0 kg block is connected to a 30.0 kg block by a string that passes over a light, frictionless pulley. The 30.0 kg block is connected to a spring that has negligible mass and a force constant of 280 N/m, as shown in the figure below. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0 kg block is pulled 16.0 cm down the incline (so that the 30.0 kg block is 36.0 cm above the floor) and released from rest. Find the speed of each block when the 30.0 kg block is 20.0 cm above the floor (that is, when the spring is unstretched).



9. A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. The force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point B, the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The speed of the block at the bottom of the track is vB = 13.5 m/s, and the block experiences an average frictional force of 7.00 N while sliding up the track.
(a) What is x?
(b) What speed do you predict for the block at the top of the track?
(c)Does the block actually reach the top of the track, or does it fall off before reaching the top?



10. A roller-coaster is released from rest at the top of the first rise and then moves freely with negligible friction. The roller coaster shown in the figure below has a circular loop of radius R in a vertical plane.
(a) Suppose first that the car barely makes it around the loop; at the top of the loop, the riders are upside down and feel weightless. Find the required height of the release point above the bottom of the loop in terms of R. (Use R as necessary.)
(b) Now assume that the release point is at or above the minimum required height. Show that the normal force on the car at the bottom of the loop exceeds the normal force at the top of the loop by six times the weight of the car. The normal force on each rider follows the same rule. Such a large normal force is dangerous and very uncomfortable for the riders. Roller coasters are therefore not built with circular loops in vertical planes. The figure and the photograph show two actual designs.



If you could show me how to do any of these that would be so amazing. This is the hardest class I have ever taken and I am a freshman at Auburn University. This is not an honors class, it is only Engineering Physics 1. I am trying to become a software engineer and physics is not what I am good at, Java is. If I fail a class I will have to go back home and work for 2 years to afford my next years tuition. I am en route to failing this class due to extreme difficulty. My high school did not make me take physics so I honestly have no idea what is going on, and webassign is making my life a living hell. Please Help me. Thanks so much for your time!

 

[kamerling]

You still have to try the problems yourself first. If you do not know what to do wil all these problems, try to study your textbook again and maybe ask questions about that.

I wouldn't combine so many problems in one question.

I can't see any of the figures, making 1 and 2 impossible

 

[Moetasim]

Dear student,

I understand that you are struggling in your Engineering Physics class and are feeling overwhelmed. I am sorry to hear that your teacher is not able to effectively convey the material to you and that you are having a hard time understanding the concepts. I am here to help you understand the content and hopefully improve your grades in the class.

First of all, let me assure you that physics is a challenging subject for many students and it is completely normal to feel overwhelmed and struggle with it. However, with the right approach and practice, you can improve your understanding and do well in the class.

Now, let's address the specific questions you have provided.

1. Conservation of Energy Equations:
(a) The bead's speed at point A can be determined using the conservation of energy equation: W_net = K_final - K_initial. In this case, the initial kinetic energy (K_initial) is zero as the bead is released from rest. Therefore, the net work done by the gravitational force (W_net) is equal to the final kinetic energy (K_final). We can calculate W_net using the work-energy theorem: W_net = mgh, where m is the mass of the bead, g is the acceleration due to gravity and h is the height from which the bead is released. Therefore, we can write the equation as: mgh = (1/2)mv^2, where v is the speed of the bead at point A. Solving for v, we get: v = √(2gh) = √(2gR), since h = 4R. Therefore, the speed of the bead at point A is v = √(2gR).

(b) The normal force on the bead can be determined using the equation: F_net = ma, where F_net is the net force acting on the bead, m is the mass of the bead and a is its acceleration. In this case, the net force is equal to the centripetal force, which is provided by the normal force. Therefore, we can write: F_net = F_c = mv^2/R. Solving for the normal force, we get: F_N = mv^2/R = m(2gR)/R = 2mg. Substituting the given mass of 5.10 g, we get: F_N = (5.10 g)(2g) = 10.2 g.

2. Particle on a friction