FL 5 - Chem/Phys Flashcards
Which of the following refers to the mechanism of peptide bond formation?
I. Nucleophilic substitution
II. Condensation
III. Hydrolysis
I and II only. Peptide bond formation occurs as the lone pair of electrons on the nitrogen atom on the amine attacks, and the amine adds to the carbonyl carbon, making it a nucleophilic substitution (I). In the process of bond formation, the hydroxyl group of the carboxylic acid leaves as a water molecule. Therefore, this is a condensation reaction (II).
What is Dynein?
Dynein is a motor protein that walks towards the minus end of the microtubules, which is oriented towards the center of the cell. Movement toward the center of the cell is described as retrograde.
Energy is measured in joules (J). What are the SI units of a joule?
Energy is measured in joules (J). The SI units for a joule are kilograms•(meters2/seconds2).
If a 1 ng red blood cell has 5 x 10-13 J of kinetic energy, how fast is it travelling in the circulatory system?
Kinetic energy = (0.5)(1 x 10-12 kg)(v2). In this case, we are given the kinetic energy of the red blood cell and asked to find the velocity, so we can rearrange the kinetic energy equation as follows:
5 x 10-13 J / [(0.5)(1 x 10-12 kg)] = v2
5 x 10-13 J / (5 x 10-13 kg) = v2
1 J / kg = v2
Taking the square root of both sides to isolate “v” gives us 1 m/s. In scientific notation this would be written as 1 x 100 m/s because 100 = 1.
What is the gravitational potential energy and the kinetic energy of a 5 g stone that is three-quarters of the way down a 7000 cm drop off a cliff in an ideal system?
Originally the stone had all of its energy stored as gravitational potential energy.
PE = mgh = (5 x 10-3 kg)(10 m/s2)(70 m) = 3.5 J.
Remember to convert 5 g into kilograms and 7000 cm into meters before plugging it into the equation for gravitational potential energy. When the stone is three-quarters of the way down a 70 m drop, it will be 17.5 m off the ground and 25 % of its original 3.5 J of energy will be stored as gravitational potential energy and 75 % will be in the form of kinetic energy. This means that three-quarters of the way down the drop, the stone will have approximately 0.9 J of gravitational potential energy and approximately 2.6 J of kinetic energy. Because the question stem specifies this is an ideal system, there is no energy lost from the system to the surroundings.
What is the gravitational potential energy and the kinetic energy of a 5 g stone that is three-quarters of the way down an 80 m drop off a cliff in a non-ideal system?
Originally the stone had all of its energy stored as gravitational potential energy.
PE = mgh = (5 x 10-3 kg)(10 m/s2)(80 m) = 4 J
Remember to convert 5 g into kilograms before plugging it into the equation for gravitational potential energy. When the stone is three quarters of the way down an 80 m drop, it will be 20 m off the ground and 25% of its original 4 J of energy will be stored as gravitational potential energy and 75% will be in the form of kinetic energy. This means that three quarters of the way down the drop, the stone will have approximately 1 J of gravitational potential energy and approximately 3 J of kinetic potential energy. Because the question stem specifies this is a non-ideal system, some energy will be lost from the system due to the presence of non-conservative forces like air resistance and friction. Therefore, the correct answer should be 1 J of gravitational potential energy and slightly less than 3 J of kinetic energy because the transfer of gravitational potential energy to kinetic energy is not 100% efficient.
If a 7 kg object moving at some velocity collides with a 5 kg object, which then hits a third object, compare the amount of work done by each of the first two objects. Assume complete transfer of energy.
According to the work-energy theorem, the amount of work done by an object is equal to the change in kinetic energy.
In other words, an object’s kinetic energy will increase in direct proportion to the amount of work we put in, and it will decrease proportionally to the amount of work the object expends. The heavier object will transfer all of its kinetic energy to the lighter object, which will, in turn, transfer all of its kinetic energy to the third object. The heavier object will be moving at a slightly lower velocity than the lighter object, but the amount of total kinetic energy transferred each time remains the same.
A 1000 kg red car traveling at 25 m/s hits a stopped 2000 kg orange car. This causes the orange car to be propelled forward into a stopped 1500 kg yellow car in front of it and causes the yellow car to move forward. How fast will the orange car be moving when it hits the yellow car? Assume the system is ideal.
The question stem says the system is ideal, so we can assume 100 % efficient transfer of kinetic energy between the cars.
KE = (0.5)mv2, so KEred car = (0.5)(1000 kg)(25 m/s)2 = approximately 312 kJ.
We can use this total amount of kinetic energy to find the velocity of the orange car because we know its mass.
312,000 J = (0.5)(2000 kg)(v2), so velocity = approximately 17 m/s.
We do not need to consider the yellow car in this scenario because it will not have an effect on the velocity of the orange car.
What is the kinetic energy of a 2 kg bird that is 100 m off the ground and flying at 3 m/s?
The formula for calculating kinetic energy (KE) is KE = (1/2)mv2, where m is the object’s mass and v is the object’s velocity. Plugging in 2 kg for the mass and 3 m/s for the velocity gives us:
KE = (1/2)(2 kg)(3 m/s)2= (1/2)(2 kg)(9 m2/s2) = (1/2)(18 kg m2/s2) = 9 J.
What are 4 different types of potential energy?
Chemical
Elastic
Gravitational
Electric
What property must an object have for it to be impossible to have elastic potential energy?
It must be incompressible.
We usually think of elastic potential energy as the energy stored in a spring, but more generally it refers to energy stored in objects due to some kind of compression. This could be a spring, or it could be a rubber band that you’ve stretched out to snap someone, or the legs of a super jumpy animal like a rabbit, or even some kind of stretchy protein or biological material.
What happens to the kinetic energy of an object when its velocity is tripled?
Kinetic energy (KE) = (0.5)mv2, where m is the object’s mass and v is its velocity. Note from this equation that kinetic energy is directly proportional to mass, and that, since velocity is squared, kinetic energy increases exponentially with how fast the object is moving. If the velocity of the object increases by a factor of 3,the kinetic energy will increase by a factor of 9 because 32 = 9.
True or false: The law of conservation of energy states that equal amounts of energy that are created must eventually be destroyed so that the total amount of energy in the universe remains constant.
This statement is false.
The law of conservation of energy states that energy is neither created nor destroyed, only transformed to different forms (i.e. kinetic, gravitational potential, elastic potential, electric, heat, etc). This law of conservation of energy is the first law of thermodynamics.
What is the gravitational potential energy and the kinetic energy of a 1 g stone that is halfway down an 800 m drop off a cliff in an ideal system? (Assume that the stone has zero gravitational potential energy at the bottom of the cliff.)
Originally the stone had all of its energy stored as gravitational potential energy.
PE = mgh = (1 x 10-3 kg)(10 m/s2)(800 m)= 8 J
Remember to convert 1 g into kilograms before plugging it into the equation for gravitational potential energy. When the stone is halfway down an 800 m drop, it will be 400 m off the ground and half of its original 8 J of energy will be stored as gravitational potential energy and the other half will be in the form of kinetic energy. This means that halfway down the drop, the stone will have 4 J of gravitational potential energy and 4 J of kinetic potential energy. Because the question stem specifies this is an ideal system, there is no energy lost from the system to the surroundings.
How much kinetic energy can a person gain if they are attached to a bungee cord that has been horizontally stretched 3 m and they are suddenly released? Assume the stiffness of the bungee cord is 5 N/m.
Elastic potential energy = (0.5)kx2, where k is the spring constant, which is an indicator of stiffness that’s unique to each spring, and x is the distance by which the spring (or bungee cord in this case) is compressed or extended. The maximum amount of kinetic energy this person could gain is equal to the amount of elastic potential energy stored in the bungee.
Elastic potential energy = (0.5)(5 N/m)(3 m)2= (0.5)(5 N/m)(9 m2) = (0.5)(45 J)
One half of 45 J is between 22.5 J.
A 6 kg object, in a non-ideal system, is initially at rest and then falls from a cliff with a height of 250 m. What is the kinetic energy of that object shortly before impact?
Because energy is always conserved, the final energy of the system will be equal to the initial energy of the system. The initial energy of the system is stored as gravitational potential energy, which we can calculate using the following formula:
PE = mgh= (6 kg)(10 m/s2)(250 m)= 15000 J
However, we need to account for the fact that some of this energy will be lost as heat due to non-conservative forces like friction and air resistance. Because this object is falling, it will experience air resistance. We can organize these concepts into the following formula:
initial gravitational potential energy = final kinetic energy + energy lost as heat
15000 J = final kinetic energy + energy lost as heat
15000 J – energy lost as heat = final kinetic energy
This means that the final kinetic energy must be some value less than 15000 J when we account for the energy that is lost as heat.
A driver in a 2,000 kg SUV moving at 15 m/s crashes into a stopped car, causing it to be propelled forward and the truck to be slowed to 10 m/s. What is the work done on the SUV by the stopped car?
We are asked to solve for the amount of work done, and we are given information about one car’s mass and velocity. We can use this to solve for initial kinetic energy of the SUV and final kinetic energy of the SUV:
KEinitial = (0.5)(2 x 103 kg)(15 m/s)2 = (0.5)(2 x 103 kg)(225 m2/s2) = 2.25 x 105 J
KEfinal = (0.5)(2 x 103 kg)(10 m/s)2 = (0.5)(2 x 103 kg)(100 m2/s2) = 1 x 105 J
There are two things to be careful about here. First, it’s important to always remember to subtract INITIAL energy from FINAL energy so that the sign is positive when energy is gained, and the sign is negative when energy is lost by the object we’re considering. Second, we need to know which object we’re talking about. The question stem specifies “work done on the SUV by the stopped car”. Intuitively we know that, because the SUV lost energy (KEinitial > KEfinal), the work done on it must be negative. We can back this up mathematically with the following:
Work done = KEfinal - KEinitial
Plugging in our values for final and initial kinetic energy gives us:
(1 x 105 J) - (2.25 x 105 J) = -1.25 x 105 J
We can assume this is an ideal system with no work lost and the only type of energy at play is kinetic energy unless otherwise indicated.
An aide is rigging a pulley system to elevate a patient’s leg. The leg weighs 12 kg and the aide needs to raise it by 0.6 m. How much work does the aide need to put into the system in order to elevate the patient’s leg?
The patient’s leg is starting off motionless in the bed with 0 J of energy and is going to be raised 0.6 m, giving it gravitational potential energy (PE). We can calculate:
PEfinal = mgh = (12 kg)(10 m/s2)(0.6 m)= 72 J
It’s important to always remember to subtract INITIAL energy from FINAL energy so that the sign is positive when energy is gained, and the sign is negative when energy is lost by the object we’re considering.
PEfinal - PEinital = 72 J - 0 J = 72 J
The aide must supply 72 J of energy to elevate the patient’s leg 0.6 m. Because energy is being added to the system (comprised of the patient’s leg), we know the final work value should be positive according to the sign conventions for work used in physics.
You may be wondering why we modeled the leg as a uniform block undergoing translational motion, rather than an object with a moment of inertia rotating about a pivot the way a real leg might. The answer is simple: always solve problems within the scope of the MCAT. As well as practicing the work-energy theorem, helping you to develop such an MCAT-question answering skill is the lesson of this practice question.
True or false: The work-energy theorem represents an exception to the law of conservation of energy because, in the real world, energy transfer between systems is not 100% efficient.
False.
The work-energy theorem is NOT an exception to the law of conservation of energy. The law of conservation is interested in the energy of an entire isolated system - that system, at least in a theoretical sense, can be as small as just the bowling ball and pins, or as big as the entire universe. The point is that individual objects in that system can gain or lose energy, so energy can be transferred from one object to another, and energy can also be transformed from one type to another. But all of that energy still stays within the system in some capacity or another. This same rule applies to scenarios when we say energy is “lost as heat”. The energy is not technically lost because if we were able to measure the amount of heat energy that was added to the system, it would account for the amount of kinetic energy or other types of energy that was lost. We only use the terminology “lost as heat” because for real-world purposes, the energy that does not do work isn’t all that useful to us.
If a red blood cell has 5 x 10-13J of kinetic energy and is moving at 1m/s, what is the mass of the cell?
Kinetic energy = (0.5)(m)(1 m/s)2. In this case, we are given the kinetic energy of the red blood cell and asked to find the mass so we can rearrange the kinetic energy equation as follows:
5 x 10-13 J / [(0.5)(1 m/s)2] = m
5 x 10-13 J / (0.5 m2/ s2) = m
1 x 10-12 kg = m
1 x 10-12 kg = 1 x 10-9 g
What is the gravitational potential energy and the kinetic energy of a 5 g stone that is one-quarter of the way down a 7000 cm drop off a cliff in an ideal system?
Originally the stone had all of its energy stored as gravitational potential energy.
PE = mgh = (5 x 10-3 kg)(10 m/s2)(70 m) = 3.5 J
Remember to convert 5 g into kilograms and 7000 cm into meters before plugging it into the equation for gravitational potential energy. When the stone is one-quarter of the way down a 70 m drop, it will be 52.5 m off the ground and 75 % of its original 3.5 J of energy will be stored as gravitational potential energy and 25 % will be in the form of kinetic energy. This means that one-quarter of the way down the drop the stone will have approximately 2.6 J of gravitational potential energy and approximately 0.9 J of kinetic energy. Because the question stem specifies this is an ideal system, there is no energy lost from the system to the surroundings.
What is the gravitational potential energy and the kinetic energy of a 5 g stone that is one-quarter of the way down an 80 m drop off a cliff in a non-ideal system?
Originally the stone had all of its energy stored as gravitational potential energy.
PE = mgh = (5 x 10-3 kg)(10 m/s2)(80 m) = 4 J
Remember to convert 5 g into kilograms before plugging it into the equation for gravitational potential energy. When the stone is one quarter of the way down an 80 m drop, it will be 60 m off the ground and 75% of its original 4 J of energy will be stored as gravitational potential energy and 25% will be in the form of kinetic energy. This means that one quarter of the way down the drop, the stone will have approximately 3 J of gravitational potential energy and approximately 1 J of kinetic energy. Because the question stem specifies this is a non-ideal system, some energy will be lost from the system due to the presence of non-conservative forces like air resistance and friction. Therefore, the correct answer should be 3 J of gravitational potential energy and slightly less than 1 J of kinetic energy because the transfer of gravitational potential energy to kinetic energy is not 100% efficient.
If a 7 kg object and a 5 kg object are each pushed until they have the same amount of kinetic energy, which object had more work done on it?
According to the work-energy theorem, the amount of work done by an object is equal to the change in its kinetic energy.
In other words, an object’s kinetic energy will increase in direct proportion to the amount of work put in, and it will decrease proportionally to the amount of work the object expends. The two objects here will move at different velocities since they have different masses, but the amount of total kinetic energy transferred to each of them remains the same.
True or false: An object in motion stays in motion if there are no forces acting on it.
This statement is true. An object in motion stays in motion if no forces are acting on it. An outside force must compel the object to stop or change its course of motion.