of Sliding and rolling: rolling ball physics
... In general, students raise two questions (i) how can the frictional force disappear?, (ii) if the frictional force disappears, what causes the torque providing the rotation? The first question derives from a poor understanding of the concept of frictional force; in this respect we must remember that ...
... In general, students raise two questions (i) how can the frictional force disappear?, (ii) if the frictional force disappears, what causes the torque providing the rotation? The first question derives from a poor understanding of the concept of frictional force; in this respect we must remember that ...
Work_Energy TN
... The work done will equal the final kinetic energy. Data Sketch a graph of velocity versus time and a graph of force versus position for one run of data. Include units and labels for your axes. (See Sample Data.) Data values in table below will be variable, due to carts having slightly different mass ...
... The work done will equal the final kinetic energy. Data Sketch a graph of velocity versus time and a graph of force versus position for one run of data. Include units and labels for your axes. (See Sample Data.) Data values in table below will be variable, due to carts having slightly different mass ...
Name - Mrs. Henderson`s Science Site
... 2. In which situation is more power required: slowly lifting a book bag full of books up the stairs or quickly lifting the same book bag full of books the same stairs? 3. Can an object at rest have energy? What type(s)? 4. How much work has a 20 N object done if it is being lifted 3 meters above the ...
... 2. In which situation is more power required: slowly lifting a book bag full of books up the stairs or quickly lifting the same book bag full of books the same stairs? 3. Can an object at rest have energy? What type(s)? 4. How much work has a 20 N object done if it is being lifted 3 meters above the ...
Document
... where the vector g = 9.8 m/s2 in the downward direction, and F = m g. ELECTRIC FIELD is obtained in a similar way: F = k q1 q2/r2 = q1 (k q2/r2) = q1 (E) where the vector E is the electric field caused by q2. The direction of the E field is determined by the direction of the F, or as you noticed in ...
... where the vector g = 9.8 m/s2 in the downward direction, and F = m g. ELECTRIC FIELD is obtained in a similar way: F = k q1 q2/r2 = q1 (k q2/r2) = q1 (E) where the vector E is the electric field caused by q2. The direction of the E field is determined by the direction of the F, or as you noticed in ...
4 Newton`s Laws
... Part I and II: Newton's Second Law is written mathematically as F ma . From this equation we see that two things affect the acceleration of a cart: the applied force and its mass. For example, a more massive cart will require a greater force in order to achieve the same acceleration as a less ma ...
... Part I and II: Newton's Second Law is written mathematically as F ma . From this equation we see that two things affect the acceleration of a cart: the applied force and its mass. For example, a more massive cart will require a greater force in order to achieve the same acceleration as a less ma ...
Chapter 12 Forces and Motion
... object’s state of motion (second law). 5. The heavier steel ball will take longer to reach terminal velocity. A greater speed is needed to produce the air resistance required to balance the steel ball’s greater weight. The steel ball must fall for a longer period of time in order to reach this gre ...
... object’s state of motion (second law). 5. The heavier steel ball will take longer to reach terminal velocity. A greater speed is needed to produce the air resistance required to balance the steel ball’s greater weight. The steel ball must fall for a longer period of time in order to reach this gre ...
IP - Uplift Education
... A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185 N horizontally. The box has a mass of 35.0 kg, and there is a force due to friction between the box and the floor that opposes motion by 25 N. Find the acceleration of the box. ...
... A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185 N horizontally. The box has a mass of 35.0 kg, and there is a force due to friction between the box and the floor that opposes motion by 25 N. Find the acceleration of the box. ...
4.1_simple_harmonic_motion_-_worksheet_
... 10. A particle of mass 0.50 kg undergoes SHM with angular frequency ω = 9.0 s-1 and amplitude 3.0 cm. For this particle, determine: (a) the maximum velocity (b) the velocity and acceleration when the particle has displacement 1.5 cm and moves towards the equilibrium position from its initial positio ...
... 10. A particle of mass 0.50 kg undergoes SHM with angular frequency ω = 9.0 s-1 and amplitude 3.0 cm. For this particle, determine: (a) the maximum velocity (b) the velocity and acceleration when the particle has displacement 1.5 cm and moves towards the equilibrium position from its initial positio ...
AP Physics Free Response Practice – Torque – ANSWERS
... extended spring gives mg = k∆x from which k can be determined. Example 2: Set the hanging mass into oscillation. Determine the period T by timing n oscillations and dividing that time by n. The equation T = 2π√ m / k can then be used to find k. b) The spring is stretched less when the object is at r ...
... extended spring gives mg = k∆x from which k can be determined. Example 2: Set the hanging mass into oscillation. Determine the period T by timing n oscillations and dividing that time by n. The equation T = 2π√ m / k can then be used to find k. b) The spring is stretched less when the object is at r ...
4.2 Fluid Friction Notes
... speed increases until at some point the upward drag equals the weight. At this point, the forces acting on the falling object are balanced and the object no longer accelerates; the speed becomes constant. The terminal speed of a falling object is the constant speed that occurs when the drag forc ...
... speed increases until at some point the upward drag equals the weight. At this point, the forces acting on the falling object are balanced and the object no longer accelerates; the speed becomes constant. The terminal speed of a falling object is the constant speed that occurs when the drag forc ...
Powerpoint Slide
... When is the velocity maximum? V = - xMAX sin(t) velocity is max. when sin(t) is max (i.e. equals 1), this happens when (t) = /2 What are we doing today? 2 experiments. The first will allow us to measure the spring constant, k, of our spring. You will hang the spring, measure the equilibrium l ...
... When is the velocity maximum? V = - xMAX sin(t) velocity is max. when sin(t) is max (i.e. equals 1), this happens when (t) = /2 What are we doing today? 2 experiments. The first will allow us to measure the spring constant, k, of our spring. You will hang the spring, measure the equilibrium l ...
PHY–302 K. Solutions for Problem set # 8. Textbook problem 7.16
... Thus, while the man walks 12 feet from the stern to the prow, the boat moves back 8 feet (i.e., 8 feet away from the pier). And by the way, the man’s displacement relatively to the pier is only 12 ft − 8 ft = +4 ft. Textbook problem 8.12: (a) For constant angular acceleration α, the angular velocity ...
... Thus, while the man walks 12 feet from the stern to the prow, the boat moves back 8 feet (i.e., 8 feet away from the pier). And by the way, the man’s displacement relatively to the pier is only 12 ft − 8 ft = +4 ft. Textbook problem 8.12: (a) For constant angular acceleration α, the angular velocity ...
Rolling Motion: • A motion that is a combination of rotational
... An ice skater would start spinning with their arms extended away from the center of her body (the axis of rotation). As the ice skater pulls her arms tight to her body, the mass is now closer to the axis of rotation, therefore the moment of inertia has been reduced and the skater spins faster in ord ...
... An ice skater would start spinning with their arms extended away from the center of her body (the axis of rotation). As the ice skater pulls her arms tight to her body, the mass is now closer to the axis of rotation, therefore the moment of inertia has been reduced and the skater spins faster in ord ...