APPLYING NEWTON’S LAWS 5.1. 5 IDENTIFY: a = 0 for each object. Apply ΣFy = ma y to each weight and to the pulley. SET UP: Take + y upward. The pulley has negligible mass. Let Tr be the tension in the rope and let Tc be the tension in the chain. EXECUTE: (a) The free-body diagram for each weight is the same and is given in Figure 5.1a. ΣFy = ma y gives Tr = w = 25.0 N. (b) The free-body diagram for the pulley is given in Figure 5.1b. Tc = 2Tr = 50.0 N. EVALUATE: The tension is the
Premium Friction Force
on the street with less than maximum static sliding friction. If the driver brakes very hard it can occur that the maximum static friction is surpassed and the wheels lose their grip and begin sliding The amount of traction which can be obtained for an auto tire is determined by the coefficient of static friction between the tire and the road. If the wheel is locked and sliding‚ the force of friction is determined by the coefficient of kinetic friction. A tire that is just on the verge
Premium
A block if wood of mass 6.0 kg slides along a skating rink at 12.5 m/s [W]. The block slides onto a rough section of ice that exerts a 30 N force of friction on the block of wood. Calculate: a. the acceleration of the block of wood. [5.0 m/s2 [E]] b. the time it takes the block of wood to stop [2.5 m/s] c. how far the block slides after friction begins to act on it. [15.6 m] 5. A force of 36 N gives a mass m1 an acceleration of 4.0 m/s2. The same force gives a mass m2 an acceleration of 12 m/s2
Free Force Mass Classical mechanics
A car is parked on a hill. In order to keep the car from rolling downhill‚ how great must the static friction acting on the car be? (Hint: Picture each scenario in your mind) equal to the force pulling the car downhill The object is experiencing some kind of friction A force is continuously applied to an object‚ causing it to accelerate. After a period of time‚ however‚ the object stops accelerating. What conclusion can be drawn? A tug-of-war that results in one team pulling the other across
Free Force Classical mechanics Mass
V. Analysis A. Vertical displacemen t= (1/2) x (Vertical Acceleration) x (Time)2 0.92m = (1/2) x (9.8m/s2) x (Time)2 Time = ((2 x 0.92m)/(9.8m/s2))1/2 = 0.43s Horizontal displacement = (Initial horizontal velocity) x (Time) 0.43m = (Initial horizontal velocity) x (0.43s) Initial horizontal velocity = Initial velocity = (0.43m/0.43s) = 1.0m/s Initial Momentum = (Mass) x (Initial Velocity) P0 = (0.008kg) x (1.0m/s) = 0.008kgm/s Time =((2 x Displacement)/(Acceleration))1/2
Premium Velocity Classical mechanics
PM Page 519 CHAPTER BERNOULLI AND ENERGY E Q U AT I O N S his chapter deals with two equations commonly used in fluid mechanics: the Bernoulli equation and the energy equation. The Bernoulli equation is concerned with the conservation of kinetic‚ potential‚ and flow energies of a fluid stream‚ and their conversion to each other in regions of flow where net viscous forces are negligible‚ and where other restrictive conditions apply. The energy equation is a statement of the conservation
Premium Fluid dynamics
magnetic field around it and the rotating armature gets attracted towards the coil. As a result the inner and outer friction plates placed between the armature and the coil gets squeezed‚ which develops a torque and eventually the vehicle comes to rest. Further researches and studies on the braking system shows that in future‚ electromagnetic braking system can be used in tandem with KERS (Kinetic Energy Recovery System). Keywords: ABS‚ EBD‚ ESC‚ Traction control‚ Emergency brake assist Brake-bywire‚ Coil
Premium
Physics Exam Study Notes By: Faizan Dhalla | SPH3U 1: Forces and Motion * Kinematics: study of motion * Basic types of motion: * Uniform (constant speed in straight line) * Non-uniform * Scalar quantity: with magnitude & no direction * Distance‚ average speed * Vector quantity: with magnitude (arrow above it) & direction (in square brackets after unit) * Position‚ displacement‚ acceleration * Velocity: * Rate of change of position *
Premium Magnetic field Electromagnetism Force
Daniel Cho Period 7 4/7/15 Electricity and Magnetism Spring Lab Project Electric Field Hockey Objective: Write two paragraphs. Explain the challenges of scoring goals in this game. Be sure to include in your explanation the terms positive and negative charge‚ electric field line and explain how mass affects the motion of the “puck.” This simulation project is all about getting the puck into the goal while surround the puck with using few positive and electric charges. The positive charges
Premium Electric charge
POTENTIAL ENERGY AND ENERGY CONSERVATION 7 7.1. IDENTIFY: U grav = mgy so ΔU grav = mg ( y2 − y1 ) SET UP: + y is upward. EXECUTE: (a) ΔU = (75 kg)(9.80 m/s 2 )(2400 m − 1500 m) = +6.6 × 105 J (b) ΔU = (75 kg)(9.80 m/s2 )(1350 m − 2400 m) = −7.7 × 105 J EVALUATE: U grav increases when the altitude of the object increases. 7.2. IDENTIFY: The change in height of a jumper causes a change in their potential energy. SET UP: Use ΔU grav = mg ( yf − yi ). EXECUTE: ΔU grav = (72 kg)(9.80 m/s2
Free Energy Potential energy Force