Homework Answers pg 98-101
... A force is needed to bounce the ball back up, because the ball changes direction, so the ball accelerates. If the ball accelerates, there must be a force. ...
... A force is needed to bounce the ball back up, because the ball changes direction, so the ball accelerates. If the ball accelerates, there must be a force. ...
Student Workbook
... 5. A 75-kg skier rides a 2830-m long chairlift to the top of a mountain. The lift makes an angle of 14.6° with the horizontal. During the accent determine: a) the change in the person’s potential energy b) the work done by the lift c) the work done by gravity. ...
... 5. A 75-kg skier rides a 2830-m long chairlift to the top of a mountain. The lift makes an angle of 14.6° with the horizontal. During the accent determine: a) the change in the person’s potential energy b) the work done by the lift c) the work done by gravity. ...
Newton`s Second Law
... Static -opposes motion before it moves (ms) • Generally greater than kinetic friction • Fmax = Force needed to get an objct moving Fmax = msFN Kinetic - opposes motion while it moves (mk) • Generally less than static friction Ffr = mkFN ...
... Static -opposes motion before it moves (ms) • Generally greater than kinetic friction • Fmax = Force needed to get an objct moving Fmax = msFN Kinetic - opposes motion while it moves (mk) • Generally less than static friction Ffr = mkFN ...
Part23 - FacStaff Home Page for CBU
... • we have Fgravity = W = mg directed down, and with air resistance (AR) • we have FAR = bv2 directed up. Newton’s Second Law gives (for 1-D): -mg + bv2 = ma , or -mg + bv2 = dv/dt which is a differential equation with v(t) being the solution. ...
... • we have Fgravity = W = mg directed down, and with air resistance (AR) • we have FAR = bv2 directed up. Newton’s Second Law gives (for 1-D): -mg + bv2 = ma , or -mg + bv2 = dv/dt which is a differential equation with v(t) being the solution. ...
Notes - Net Forces and Applications of Newton`s Laws
... move. The force of static friction exerted by the surface on the object prevents it from moving. The object does not move until you push hard enough (Fmax)to overcome static friction. Fmax = sN where s is the coefficient of static friction. ...
... move. The force of static friction exerted by the surface on the object prevents it from moving. The object does not move until you push hard enough (Fmax)to overcome static friction. Fmax = sN where s is the coefficient of static friction. ...
Problem Solving—A General Approach | Summary
... 12. You pull a box with a constant force across a frictionless table using an attached rope held horizontally. If you now pull the rope with the same force at an angle to the hori zontal (with the box remaining flat on the table), does the acceleration of the box (a) remain the same, (b) increase, o ...
... 12. You pull a box with a constant force across a frictionless table using an attached rope held horizontally. If you now pull the rope with the same force at an angle to the hori zontal (with the box remaining flat on the table), does the acceleration of the box (a) remain the same, (b) increase, o ...
Practice Test.100A 4-5
... In (a) the friction is static and opposes the impending motion down the plane. In (b) the friction is kinetic and opposes the motion down the plane. In (c) the friction is kinetic and opposes the motion up the plane. 6. A box given a push so that it slides across the floor. How far it will go given ...
... In (a) the friction is static and opposes the impending motion down the plane. In (b) the friction is kinetic and opposes the motion down the plane. In (c) the friction is kinetic and opposes the motion up the plane. 6. A box given a push so that it slides across the floor. How far it will go given ...
Physics I - Rose
... Let us obtain n from the y-component equation as n mg cos, and substitute it in the above equation to get Fpush mg sin k mg cos mg (sin k cos) (100 kg)(9.80 m/s2)(sin 20 0.60 cos20) 888 N The force is less than your maximum pushing force of 1000 N. That is, once in motion, ...
... Let us obtain n from the y-component equation as n mg cos, and substitute it in the above equation to get Fpush mg sin k mg cos mg (sin k cos) (100 kg)(9.80 m/s2)(sin 20 0.60 cos20) 888 N The force is less than your maximum pushing force of 1000 N. That is, once in motion, ...
C) Activities/Resources for Module Outcomes
... Force is an influence or energy that causes an object to undergo a change or an interaction with another object. Friction is the force between two objects as they move over one another, such as a vehicle’s tire and the surface it is traveling on. There are two types of friction: static and kinetic. ...
... Force is an influence or energy that causes an object to undergo a change or an interaction with another object. Friction is the force between two objects as they move over one another, such as a vehicle’s tire and the surface it is traveling on. There are two types of friction: static and kinetic. ...
7th gd Forces
... Book M – Ch 2 Section 3 (pgs. 51-54) Newton’s First Law of Motion • States that an object at rest will remain at rest, and an object moving at a constant velocity will continue moving at a constant velocity, unless it is acted upon by an unbalanced force. • Inertia - the tendency of an object to res ...
... Book M – Ch 2 Section 3 (pgs. 51-54) Newton’s First Law of Motion • States that an object at rest will remain at rest, and an object moving at a constant velocity will continue moving at a constant velocity, unless it is acted upon by an unbalanced force. • Inertia - the tendency of an object to res ...
05._UsingNewtonsLaws
... T1 T1 cos , sin T2 T2 cos , sin x : T1 cos T2 cos 0 ...
... T1 T1 cos , sin T2 T2 cos , sin x : T1 cos T2 cos 0 ...
Solutions to Assignment #1
... Now calculate the horizontal distance traveled. s x = vx t = 2.047(0.3954) = 0.8095 m Thus, she will travel 0.810 m or 81.0 cm horizontally. #4. A long jumper takes off with a velocity of 11.00 m/s at an angle of 25.0o. The athlete’s centre of gravity was 1.250 m above the ground at takeoff and 0.50 ...
... Now calculate the horizontal distance traveled. s x = vx t = 2.047(0.3954) = 0.8095 m Thus, she will travel 0.810 m or 81.0 cm horizontally. #4. A long jumper takes off with a velocity of 11.00 m/s at an angle of 25.0o. The athlete’s centre of gravity was 1.250 m above the ground at takeoff and 0.50 ...
QUICK QUIZZES 1. Newton`s second law says that the acceleration
... force acting on. Recognizing this, consider the given statements one at a time: (a) True— If the resultant force on an object is zero (either because no forces are present or the vector sum of the forces present is zero), the object can still move with constant velocity. (b) False — An object that r ...
... force acting on. Recognizing this, consider the given statements one at a time: (a) True— If the resultant force on an object is zero (either because no forces are present or the vector sum of the forces present is zero), the object can still move with constant velocity. (b) False — An object that r ...
2 Friction and Gravity
... Static Friction Four types of friction are shown in Figure 6. The friction that acts on objects that are not moving is called static friction. Because of static friction, you must use extra force to start the motion of stationary objects. For example, think about what happens when you try to push a ...
... Static Friction Four types of friction are shown in Figure 6. The friction that acts on objects that are not moving is called static friction. Because of static friction, you must use extra force to start the motion of stationary objects. For example, think about what happens when you try to push a ...
Chapter 5-6
... force must be at its maximum on the opposite direction to the spring force. Maximum friction force = N * s = mg cos(65°) * 0.18 = 1.49N Along the incline plane, all forces projections add up to zero Therefore, Fspring – Fr – mg sin(65°) = 0 and Fspring = k x thus x = Fspring / k = (Fr + mgsin(65°))/ ...
... force must be at its maximum on the opposite direction to the spring force. Maximum friction force = N * s = mg cos(65°) * 0.18 = 1.49N Along the incline plane, all forces projections add up to zero Therefore, Fspring – Fr – mg sin(65°) = 0 and Fspring = k x thus x = Fspring / k = (Fr + mgsin(65°))/ ...
AP Physics – The Physics Little AP Test Review Helper
... A. Object has constant horizontal motion at beginning of fall. Examples: Bomb being dropped from aircraft, bird drops an object, ball rolling along table and then falling to floor, &tc. Possible problems: (1) Given height of fall and horizontal velocity find the horizontal distance that the ...
... A. Object has constant horizontal motion at beginning of fall. Examples: Bomb being dropped from aircraft, bird drops an object, ball rolling along table and then falling to floor, &tc. Possible problems: (1) Given height of fall and horizontal velocity find the horizontal distance that the ...