MCAT Princeton Review Physics

Person walks east along the road, what frictional force does he feel

-A static force east -Not kinetic force, otherwise he would be slipping (Kinetic force does not apply to all moving objects, only slipping objects) -As the person walks east, their feet push back against the road to the west. So it must be working in the east direction

Book slides across the table, what net force did the book exert on the table

-Diagonal force due to the reactive normal force and the reactive force of kinetic friction (forces exerted by the book on the table) -Solve for the vector of these forces!

Torque

IF I: Moment arm F: Force -Measured in Nm (equivalent to joules) -An object can have a force acting on it without exerting a torque if that force's line of action passes through the center of mass of the object

Conservation of Energy formula

KEi+PEi+Wi=KEf+PEf+Wf -Make sure to take in account for KEi or Wi when setting up this problem!

Moment arm

The component of r that is perpendicular to the force -Distance from the pivot to the line containing the force

Center of mass formula

[m1X1 + m2x2 + Mx (mp)] / [m1 + m2 + M] -Can use this to solve a teeter totter problem -CM is where the fulcrum should be placed

Centripetal acceleration formula

a=v^2/r

Normal force and solid molecules

-It is attributable to the binding forces between atoms and molecules in solid matter

Translational equilibrium

-Net force equals zero, so acceleration equals zero and velocity but be constant -Curved paths do no have constant velocity because direction changes

Mechanical advantage of a machine

-ratio of the force required to lift an object a given distance to the force required to move that object the same vertical distance (Fapplied/mg) -The work involved is the same whether the mass is lifted straight up or pushed up the ramp -Can solve by calculating d/h: when you set the W equations to equal each other and solved for Fapplied/mg

Can a normal force cause a moving object to change directions?

-yes, On a circular track, the normal force provides some on the centripetal acceleration

Cos(45)

0.7

Ideal projectile: displacement at T/2 and at T

2d(T/2)>d(T) -At half the total flight time, a projectile launched upward will reach its apex. All velocity is in the horizontal direction, acceleration is constant through out the flight. -d(T) is the total range -d(T/2): is the hypotenuse formed from the launch point to the apex. This is longer than half the total range!`

Power equation for constant force parallel to velocity

P=Fv