1. Net force and weight

2.  How much would a person who weights 490 N on Earth weigh on Jupiter, if Jupiter’s gravity is 23 N/kg

3.  You know that F=ma or F=mg  You know that your mass is constant  M= 490 / 9.8 = 50kg  Use the mass to plug into find the force or weight of the person with the gravity on Jupiter  F= 50*23 = 1,150 N

4.  You get on an elevator and you weight 100 lbs.  You step on a scale, when the elevator goes up, the scales says that…. 1. You weigh more than 100 lbs 2. You weigh 100 lbs 3. You weigh less than 100lbs 4. You weigh nothing

5.  You ‘weigh’ more than 100 lbs  The person’s inertia wants stay stationary, so the elevator floor has to push up on the person.  There forth the scale has to push upward with extra force on the person to accelerate the person’s mass upward

6.  You get on an elevator and you weight 100 lbs.  You step on a scale, when the elevator goes up, the cord pulling you up breaks and you start to fall, the scales says that…. 1. You weigh more than 100 lbs 2. You weigh 100 lbs 3. You weigh less than 100lbs 4. You weigh nothing

7.  Weighs nothing  You and the elevator are all falling together at the same downward acceleration, so there is no contact force (normal force)

8.  You get on an elevator and you weight 100 lbs.  You step on a scale, when the elevator goes down, the scales says that…. 1. You weigh more than 100 lbs 2. You weigh 100 lbs 3. You weigh less than 100lbs 4. You weigh nothing

9.  Why?  The inertia of the person wants to stay at rest, so the elevator floor and the scale effectively drop put a little bit from underneath the person as it goes down

10.  Calculate the total net force the diagram below

11.  First calculate the x and y components separately  Fy= +3 + -3 = 0 newtons  Fx= -5 = -5 newtons  Only has -5 N in the x direction

13.  Fx= -60 = -80 + x  X= 20 N Fy= 0 = +300 + y y= -300

14.  Suppose that a sled is accelerating at a rate of 2 m/s 2. If the net force is tripled and the mass is halved, then what is the new acceleration of the sled?

15.  Know that a= 2  Suppose m=1  F=2*1  F=2  If says f is triples and m is halved  F2= 6  (2*3)  M2= ½=,5  6=.5(a2)  A2= 6/.5 = 12 m/s^2

16.  An applied force of 50 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the mass, and the acceleration of the object. (Neglect air resistance.)

17.  F norm = 80 N; m = 8.16 kg; F net = 40 N, right; a = 4.9 m/s/s, right  ( F norm = 80 N; m = 8 kg; F net = 40 N, right; a = 5 m/s/s, right )  Since there is no vertical acceleration, normal force = gravity force.The mass can be found using the equation F grav = m g.  The F net is the vector sum of all the forces: 80 N, up plus 80 N, down equals 0 N. And 50 N, right plus 10 N, left = 40 N, right.  Finally, a = F net / m = (40 N) / (8.16 kg) = 4.9 m/s/s.