How Does a Cookie Jar's Motion on an Incline Relate to Physics Principles?

[lethalblade]

1.) A cookie jar is moving up a 39.7o incline. At a point 0.56 m from the bottom of the incline (measured along the incline), it has a speed of 1.48 m/s. The coefficient of kinetic friction between the jar and incline is 0.15.

(a) How much farther up the incline will the jar move?

(b) How fast will it be going when it slides back to the bottom of the incline?

2.) A pendulum (87cm long) with a 0.75kg bob is released from rest at an initial angle of theta with the vertical. At the bottom of the swing, the speed of the bob is 2.9m/s. Find the initial angle theta:

A pendulum (87cm long) with a 0.75kg bob is released from rest at an initial angle of theta with the vertical. At the bottom of the swing, the speed of the bob is 2.9m/s. What angle does the pendulum make with the vertical when the speed of the bob is 1.1 m/s?

Any help would be greatly appreciated.

 

[hotvette]

Please let us know what approach you are supposed to be using (e.g. F=ma, work-energy, etc.) and describe what you've tried so far and where you are stuck.

https://www.physicsforums.com/showthread.php?t=94379

 

[quicknote]

For the first problem, we can use the equations for motion on an inclined plane to solve for the distance the cookie jar will move up the incline and its speed when it reaches the bottom again. We know that the acceleration of the jar is determined by the force of gravity pulling it down the incline and the force of friction acting against it, which is given by the coefficient of kinetic friction. We can use the equation a = gsinθ - μgcosθ, where θ is the angle of the incline and μ is the coefficient of kinetic friction. We also know that the distance traveled is given by the equation x = x0 + v0t + 1/2at^2, where x0 is the initial position, v0 is the initial velocity, and t is time.

(a) To find how much farther the jar will move up the incline, we can set x0 = 0, v0 = 1.48 m/s, and solve for x when t = 0. We know that the jar will eventually come to a stop, so we can set the final velocity to 0 in the equation. This gives us x = 1.48^2/2(0.15)(9.8)sin39.7 = 0.71 m. Therefore, the jar will move 0.71 m farther up the incline before coming to a stop.

(b) To find the speed of the jar when it reaches the bottom of the incline, we can set x0 = 0, v0 = 1.48 m/s, and solve for v when x = 0.56 m. This gives us v = √(1.48^2 + 2(0.15)(9.8)(0.56)sin39.7) = 0.56 m/s. Therefore, the jar will be moving at a speed of 0.56 m/s when it reaches the bottom of the incline.

For the second problem, we can use the equation for the conservation of energy to solve for the initial angle theta. At the bottom of the swing, all of the potential energy is converted into kinetic energy, so we can set the initial potential energy equal to the final kinetic energy. This gives us mgh = 1/2mv^2, where m is the mass of the bob, g is the acceleration due to gravity, h