Physics AAMC Qpack: question 74,75,79,82,87,89, 91, 92, 95

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lotus0618
Posts: 182
Joined: Sun Dec 29, 2019 5:47 pm

Physics AAMC Qpack: question 74,75,79,82,87,89, 91, 92, 95

Post by lotus0618 » Tue Jun 16, 2020 10:21 pm

Question 79:
W = Fdcos θ and F = ma, in which a depends on speed.
Can you please explain to me why work is independent of v here?

Question 74:
Why isn’t the angle counted in the equation of W=Fdcos(theta)?

Question 75:
Why does frequency change here? For instance, question 93, frequency status constant because it’s still the same source and is independent of the medium.

Question 82:
I got this right but can you please check if my reasoning is correct?
W = − μkhNd. Here, only Fnormal is affected. The passage indicates that W is generated as the box travels down the ramp, so I chose the horizontal component of Fnormal, which is N= mg cos(theta). Greater angle results in smaller cos(theta), so W will decrease.

Question 87:
Why doesn’t The kinetic energy of a sliding block come from the gravitational potential E of the block? I know I’m missing a basic important concept here. Can you please explain it to me?

Question 89:
When the block began to slide, I reasoned that the speed would stay constant in time because there’s no other force acting on it to make it accelerate.
In fact, someone explained to me that the Force of gravity is greater than the friction force, which will accelerate the box. But the problem indicates that the block already started sliding.

Question 91: Can you please explain this?

Question 92: Can you please explain this one?

Question 95: I know the torque equation and I know how to calculate the type of problem. But i’m having a hard time understanding why 50 cm, not 30 cm is the center of mass

Thank you!
NS_Tutor_Nancy
Posts: 47
Joined: Wed May 27, 2020 7:25 pm

Re: Physics AAMC Qpack: question 74,75,79,82,87,89, 91, 92, 95

Post by NS_Tutor_Nancy » Thu Jun 18, 2020 10:33 pm

Hello!

Q74: While you are pulling the rope at an angle, looking at the pulley set up, the weight is actually pulled directly upwards, so force and displacement are in the same direction, making theta 0 degrees. Cos (0) = 1, so you can essentially just drop it out of the equation.
Q75: The observed frequency would change because there is a reflector moving toward the source. So if you were standing at the source and listening, the echo sound reflected back would have a higher frequency than the sound coming from the source.
Q79: Careful here, W is specifically defined in this passage as referring to the work done by friction, so the force in the work equation is the friction force. The friction force will equal the normal force on the mass times the coefficient of friction, it has nothing to do with velocity!
Q82: Exactly right. Also just thinking about it conceptually, increasing the angle of the ramp making it steeper would make anything slide down faster/easier as you get closer to 90 degrees, meaning friction is doing less work!
Q87: So gravitational potential energy is coming from the distance off the ground. Here the mass on the hook falling down and losing its gravitational potential energy is what causes the sliding block to slide. The sliding block is flat on a table, so it’s gravitational potential energy isn’t changing.
Q89: Since there is a non-zero net force, thinking F = ma, there is certainly going to be some acceleration. The block is already sliding, but there is still friction acting on it (kinetic rather than static friction), it is just that the force provided by gravity is overcoming friction.
Q91: Buoyant force = weight of liquid displaced = density of liquid x volume of submerged object x g. In both cases here, the object is totally immersed so the volume stays constant, and g is also a constant, so only density of liquid changes. That means that the ratio of buoyant forces will equal the ratio of liquid densities. This gives you 5/12 = 0.7/x >>> x = (12x0.7)/5 = 1.7
Q92: Think of the kinematic equation vf = vi + at. Let’s start with the upward journey, vf will be 0 when it hits the highest point and starts falling down, so you can simplify and solve for time: t = v/a. In this situation time up will equal time down, so total t = 2v/a. Here acceleration is g, so if it were reduced to g/6, 2v / (g/6) = 2v x 6/g = 12v/g. So you increased by a factor of 6.
Q95: You want the distance between the mass and the suspension point. If the suspension point is at 30 cm on the meter stick and the mass is at 80 cm, your lever arm is 50 cm.

Hope that helps!
lotus0618
Posts: 182
Joined: Sun Dec 29, 2019 5:47 pm

Re: Physics AAMC Qpack: question 74,75,79,82,87,89, 91, 92, 95

Post by lotus0618 » Fri Jun 19, 2020 3:22 pm

74: Would we consider the angle if the force caused the box to go in the opposite direction? Under what scenarios in which we don't have to consider the angle?

75: How am I supposed to know that the problem refers to the observed f? I interpreted this as the source f,so that's why I thought it didn't change before and after the reflection

91: how am I supposed to know that I have to do ratios for this question? I have come many problems in which doing ratios will help solve the problems easily, but I wasn't aware of it.

95: I still cannot visualize the problem. Can you please draw it out and show it to me?

Thank you
NS_Tutor_Yuqi
Posts: 57
Joined: Fri May 29, 2020 11:43 pm

Re: Physics AAMC Qpack: question 74,75,79,82,87,89, 91, 92, 95

Post by NS_Tutor_Yuqi » Sun Jun 21, 2020 2:36 pm

74. If the object is being moved at an angle instead of directly up, then the angle would have to be considered into the equation.

75. Since the sound wave is being reflected off the imperfect reflector, it loses energy and therefore the frequency also changes since frequency is directly proportional to energy. As for interpreting f to be the source f, the question describes how the sound wave is traveling and bouncing off the reflector. Therefore, we wouldn't want to just think about what the frequency was at the sources since we know that the sound is being manipulated by the moving reflector.

91: Whenever a question is setting up a relationship between two scenarios, you should automatically consider whether it would be useful to set up a ratio to determine the unknown variable. Although setting up ratios may not help you in every case, it will be good to develop a habit of considering the possibility so that you're prepared when ratios are needed to find the answer.

95: See attached image
Meter Stick Drawing.jpg
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