Antecedentes a nivel Regional

1. B Begin by finding the acceleration caused by the frictional force. The equation v² = v₀² + 2ad, when the final velocity is set to zero and when the equation is solved for a, yields that the magnitude of the accelera-tion is ⁰

2 a v

= d. Now apply F_net = ma: the net force is supplied by kinetic friction, so µmg = ma, and thus a = µg. Combining the two equations gives 2⁰

g v

m = d _{and then} ⁰ 2

v

m = dg . Plugging in v₀ = 4, d = 2, and g = 10 gives a result of µ = 0.4. Additional note: this problem does not require any information from the passage; it is a disguised free-standing question.

2. C The passage states that coefficients of friction are usually lower at very high speeds. Consequently, the force of kinetic friction is lower at very high speeds. Therefore, the force required to counter kinetic fric-tion is lower at very high speeds. Choice A is a trap, because the net force on an object at constant speed is zero, but there will be a frictional force slowing the object down, so it needs an additional push to keep going. Choice B is backwards. Choice D would be true if the coefficient of friction were constant at all speeds, which is the usual approximation on the MCAT, but the passage indicates that the coefficient of friction should not be considered constant here.

3. D Since the object is sliding, it is experiencing a kinetic friction force, which is equal to µF_N. Since it is not accelerating vertically, the sum of the forces in the vertical direction is zero, which means that the normal force upwards initially balances only the weight of the object downwards (that is, F_N = w). Thus, the force of kinetic friction is µw, which equals 20µ. After the additional pressing down, the normal force balances the weight and the extra downward force, so the force of kinetic friction is (µ)(20 + 40), which equals 60µ. That’s three times as large, so the new force must be 18 N, three times as large as it was before. (One could

solve for what the coefficient of friction actually is to cause this force, which is 0.3, but it is not necessary to do so in this case; proportions are sufficient here.) Additional note: this problem does not require any information from the passage; it is a disguised free-standing question.

4. A This is an inclined plane problem involving a flat tilted surface. From typical inclined plane analysis, the forces on the pot parallel to the plane are mg sin θ down the plane and static friction up the plane. Right before the pot starts to slide, static friction is at its maximum and the two forces are equal, so mg sin θ = µmg cos θ, which yields that µ = tan θ. From Table 1, the coefficient of static friction between zinc and cast iron is 0.85. Compare this to tan(45°), which is equal to 1; 0.85 is smaller, so the angle must be less than 45°; Thus, it must be tan(40°), since 40° is the only choice less than 45° available.

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5. B This is a two-by-two question: two answer choices say that the coefficient of friction is lower and two say that it is higher, and two answer choices say that the operative friction is static and the other two say that it is kinetic. First, the coefficient of friction is lower on an icy road. As the passage indicates, the coefficient of friction is usually a measure of roughness, and icy roads are smoother than non-icy roads.

Also, the question stem says that cars are more likely to slip on the ice, and rolling objects slip because the frictional force does not provide enough torque to keep the objects rolling; here, the force is reduced because the coefficient of friction drops. This eliminates choices C and D. Second, the first paragraph of the passage indicates that objects that are rolling without slipping are being moved by static friction, so static friction is the relevant type here, eliminating choice A.

6. C First, decipher the wording of the question. The variable F_x refers to the amount of force required to initi-ate horizontal movement, which means the force required to overcome static friction. The force to over-come static friction is equal to the maximum force of static friction, µF_N, which in turn is equal to µw.

The ratio that several of the answer choices talk about is F_x:w, which is µw / w, which is the same thing as µ / 1. In other words, all that this question is asking is what happens to the coefficient of friction as more weight is added to the rubber object. Now, refer to the passage: Equation 1 describes how the coefficient of friction changes for rubber when the normal force from the surface increases. In particular, the recipro-cal of the coefficient of friction increases linearly with the normal force, so the coefficient of friction must decrease with normal force; more weight means less coefficient of friction. Therefore the ratio goes down.

7. B Begin by referring to the passage. PTFE is mentioned in the table, and it has a coefficient of friction of 0.04, which is much lower than anything else in the table. The answer is likely to have to do with this low coefficient of friction. A might be tempting because it suggests that PTFE is not very toxic, but lack of toxicity itself is not a major reason why PTFE is applied to cookware; any other non-toxic substance might equally be applied if that were its only useful property. Choice B addresses the low coefficient of friction and at least is physically possible, because low coefficients of friction enable sliding more easily than high coefficients of friction, and easy sliding means less sticking. Choice C is non-physical: low fric-tion means less heat generated by fricfric-tion, not less heat lost to fricfric-tion. Choice D is off-topic; this may or may not have anything to do with cooking.

KINEMATICS AND DYNAMICS

Drill

CHAPTER 38 PRACTICE PASSAGE

Jack and Jill are running up a hill, at an incline of approximately 30° to the horizontal, along a path that is 450 m long to the top of the hill, as shown in Figure 1. Jack weighs 900 N and Jill is half his size. They are racing to the top of the hill, and Jill runs up in half the time it takes for Jack to scramble up the hill. Their good friend Ty-Mur measures how long it takes for them to get to the top of the hill and clocks in Jack at 3 minutes.

450m 30º

Figure 1 Jack and Jill’s Hill

At the top of the hill is a well where they try to fetch a pail of water. The well is 15 m deep, but Jill cannot see the bottom.

They decide to drop a penny down to check if there is water in the well. Then they lower the 2 liter pail using a pulley system into the well and draw up a full bucket. Then they start walking back down the hill.

Unfortunately, Jack falls when they are almost at the bottom of the hill when they have travelled 440 m from the summit, and he slides down the hill for 6 m before coming to a stop. Jill drops the pail and races down after him. She trips just as she gets close to him, and they wind up sliding down the hill together, coming to a stop just as they reach the bottom of the hill.

1. When Jack has reached the top of the hill, how much Jill dropped the penny and stopped it when they heard the splash from the penny she dropped?

A) 0.04 s B) 0.08 s C) 1.77 s D) 3.46 s

4. If Jill was running down the hill at 9 m/s before falling onto Jack, what fraction of her kinetic energy was lost in the collision? the trip to the end of the trip?

A) 0 J B) –101.25 kJ C) 101.25 kJ

D) Cannot be determined

6. Let W₁ be the work done by friction in stopping Jack after fall, and W₂ be the work done by friction in stopping Jack and Jill together after Jill’s fall onto Jack. Assuming the same slope and same value of µ between them and the ground in both cases, which of the following is true about the relation between these quantities?

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