1.) Answer (D) Foot applies 200 N force to nose, nose applies an equal force to the foot. Basic application of Newtons 3rd Law. 2.) Answer: (C) The basketball pushes Joanne to the west. (although the basketball applies a force to Joanne, she wont accelerate if there is another force acting on her, such as the force of friction between her and the floor, to balance out the force.) 3.) Answers: B & D: The force of air resistance on the book is greater than the force of air resistance on the feather due to the larger cross-sectional area of the book (the book moves through more air). However, the book also experiences a larger downward force due to gravity due to its larger mass. Even though the force of air resistance is greater for the book, the proportion of the force of air resistance to the force of gravity on the book is smaller than that for the feather. The book has a larger net force on it, and also a larger proportion of force to mass, resulting in a larger downward acceleration. 4.) Answer: (B) mg(1-cos) Begin with a diagram, free body diagram.

Writing a Newtons 2nd Law equation in the y-direction provides the appropriate relationship to solve for the applied force.

5.) Answer: (C) The tension in the spheres would double. When each team of horses is pulling on the spheres, the tension from each team is the same, with each team pulling in opposite directions. Whatever force one team pulls with, the other team must pull back with an equal magnitude force, otherwise the sphere would accelerate. This is the same as the force that would be exerted if one side of the sphere was held motionless while one team of horses pulls on a hemisphere. By attaching both teams (all 30 horses) to the same hemisphere, and fixing the opposing hemisphere in place, double the force of one team is obtained between the hemispheres.

6.) Answer: (A) 2T1/7 Remember, when dealing with a system, you deal with the system as a whole first! That means you consider the net force acting on the system as a whole (in this case, T1) and the entire mass of the system (in this case 7 kg) to solve for the acceleration of the whole system (a=T1/7kg) . Then, make use of the fact that all parts of a system must have the same acceleration (They are moving together). Then, you can analyze the force acting on a piece of the system, along with the mass of just that piece of the system.

(although not really necessary for this problem, remember that all internal forces (forces acting within a system) dont contribute to the net force or acceleration of the system. T2 here is an internal force!

Analyzing the system as a whole, T1=(7kg)a, therefore a=T1/7kg Looking at just the 2-kg block, T2=(2kg)a. Substituting in the acceleration of the system (since both blocks must have the same acceleration), you find T2=2T1/7

7.) If the acceleration of the system is the same, then T1 must increase in order to provide the necessary force to keep the acceleration the same. F=ma. Since the mass of the system increased, the force (T1) must also increase. T2 must increase for the same reason, since it is now acting on more mass. However, we can be more specific with our answer. Lets solve for how much force it would take to accelerate the new system in terms of T1.

Our new T1 has to be 9/7 as great as the old T1. We can do the same thing for our new T2 that would be required to accelerate 4 kg now instead of 2 kg.

Notice that this is twice the amount T2 was in the first part of the problem.

8.) T2 is pulling 4 kg forward.

9.) Draw a free-body diagram. Notice that forces acting in the x-direction (the direction parallel to the slope) are the x-component of gravity (Fgsin()) and the force of friction. The normal force is equal in magnitude to Fgcos. (If you dont remember why, look back to your notes on incline planes!!!) If Jane is going at a constant speed, that means the acceleration is zero, and the net force acting on Jane is also zero! The x-component of gravity must then be equal in magnitude to the force of friction.

10.) First draw a free body diagram for the two masses. Next you can write out Newtons 2nd Law equations for the two objects, keeping in mind the y-axis we have chosen curving around the pulley. T1 m1g = m1a m2g T2 = m2a Combining these equations and recognizing T1=T2 because this is an ideal, massless pulley, you can then solve for the acceleration of the system.

Then, you can use kinematics equations and the acceleration you just found to solve for the distance.

11.) A > D=F > B > C=E A simple analysis of the Atwood Machines shows that the acceleration of the system is equal to the net force divided by the sum of the masses. Summarizing in table form:

12.) Answer: The vehicle remains at rest. Students may approach their justification in multiple ways. An example of a strong answer could include references to all three of Newtons Laws (i.e. the vehicle will remain at rest because it is not acted upon by an outside force. You can observe this as the spinning fan imparts a force on the air molecules, which, by Newtons 3rd Law, apply an opposing force on the fan blades. The air molecules impact the sail, imparting a force on the sail, while the sail imparts an opposing force back on the air molecules. Because both the fan and the sail are part of the same object, the opposing forces on the sail and fan blades balance each other out, resulting in no net force. According to Newtons 2nd Law, if there is no net force, there is no acceleration, therefore the vehicle remains at rest.

13.) A free body diagram showing all the forces on the rope is quite helpful in this situation. Note that the tension in the portion of rope connected to the fence post must be the same as the tension in the portion of rope connected to Paisley.

If the force is just strong enough to pull the horse out of the mud, then the net force acting on the rope is zero. That means the rope must be pulling with 500 N to the right to balance Lindas 500 N to the left. However, that 500 N is coming from the x-component of the tension in the rope. To solve for the tension, note that rope is pulling Linda from two sides, the side from the fence post and the side from the horse. Each side will be offering up half of the force needed to balance out the 500 N from Linda.

14.) First draw a free body diagram for the car on the banked curve, then another diagram showing the components of the normal force acting on the car.

The normal force acts perpendicular to the surface. Since the car is going around a curve, there must be a net centripetal force acting on the car, acting in the direction of the center of the curve. From the diagram above, the only force acting in that direction is the x-component of the normal force. Therefore, it is equal to the net centripetal force. Also from the diagram, the y component of the normal force is balancing out gravity. Write Newtons 2nd law equations for the x and y directions. Then, solve for the normal force using the y direction. Substitute that into your equation for the x-direction, and solve for the angle.

15.) The students mass has inertia and wants to stay at rest. In order to accelerate her mass, an unbalanced force must be applied. The reading on the scale shows not only the force of her normal weight, but has an additional unbalanced force causing her mass to accelerate.

16.) Answer:

You must first recognize that the radius of the satellites orbit is 3R, the radius of the planet plus the altitude of the satellite above the surface of the planet. Then, a force analysis recognizing the gravitational force of attraction provides a centripetal force yields:

17.) Answer: bF=mg. Earth exerts a greater force on the object that has greater mass. (if both were dropped, they would have the same acceleration)

18.) Answer: cM is the mass of the object creating the gravitational field (Earth), not the object in it. In that case, all the variables in the formula for the gravitational field strength at that height above the Earth are the same.

19.) Answer:cThe force of gravity between two object is from the formula to the left. The objects exert the same force on each other. (This could also be thought of from a Newtons third law perspective. The forces are an action-reaction pair)

20.) this is the formula for gravitational field strength. First, see how the value for g would change if M is doubled, and R is halved.

The gravitational field strength is 8 times as strong.

If a person weighs 700 N on Earth (lets use 10 m/s2 for g) then we can calculate his/her mass.

Then we can calculate the weight with the gravitational field strength that is 8 times as strong.

21.) You must first find what happens to the tension in the rope if the speed is doubled. Tension is the force causing circular motion, so it is a centripetal force.

The tension is quadrupled. Now we must deal with the uncertainty. Multiply the smallest possible value for tension by 4, and the biggest possible value by 4.

Those are your biggest and smallest predicted values. To get your value with uncertainty, add them up and divide by two. Then, include your range as the uncertainty.

22.) Again, deal with the system as a whole first. Lets say that the direction m2 is falling is the positive direction. Then, the forces acting on the system are the force of gravity from m2 in the positive direction, and the x-component of the force of gravity acting on m1 in the negative direction. The forces of tension cancel out (They are internal forces!) We can then use newtons 2nd law for the system to find the acceleration of the system.

To find the tension, we will analyze the block m2, and consider the two forces acting on it. (You could also use m1 instead. Try it! see if you get the same answer!) m2 must have the same acceleration as the system. The net force acting on it will be the force of gravity acting on it minus the force of tension.

23.) As the applied force is increased, the force of static friction must increase as well to keep the box in equilibrium. However, once the applied force is greater than the maximum force of static friction, the box will start to move. Now the applied force is kept the same. However, since the box is moving, it is now resisted by kinetic friction. The magnitude of kinetic friction is lower than the maximum force of static friction. Therefore, the applied force is greater than the force of friction, and there is a net force in the direction of motion, meaning that the box will accelerate.