Review Questions on Jeat Transfer and Conservation Energy

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All MCAT Physical Resources

A ball with mass of 2kg is dropped from the acme of a edifice this is 30m high. What is the judge velocity of the brawl when it is 10m higher up the ground?

Possible Answers:

20m/s

30m/s

25m/south

14m/s

10m/s

Explanation:

Use conservation of energy. The gravitational potential energy lost as the ball drops from 30m to 10m equals the kinetic free energy gained.

Change in gravitational potential energy tin be found using the divergence in mgh. $mgh_{f} - mgh_{i} = 2kg*10\frac{m}{s^{2}}*10m - 2kg*10\frac{m}{s^{2}}*30m = -400 Joules$

So 400 Joules are converted from gravitational potential to kinetic energy, allowing us to solve for the velocity, v.

$400 Joules = \frac{1}{2}mv^{2}$

$400 = v^{2}$

$v = 20 m/s$

Consider a spring undergoing simple harmonic motion. When the spring is at its maximum velocity __________.

Possible Answers:

kinetic energy and potential free energy are at a maximum

kinetic energy is at a minimum and potential energy is at a maximum

kinetic free energy is at a maximum and potential free energy is at a minimum

kinetic energy and potential free energy are at a minimum

Right answer:

kinetic free energy is at a maximum and potential free energy is at a minimum

Explanation:

Kinetic energy is highest when the spring is moving the fastest. Conversely, potential energy is highest when the spring is most compressed, and momentarily stationary. When the force resulting from the compression causes the spring to extend, potential energy decreases as velocity increases.

A pendulum with a mass of 405kg reaches a maximum height of 2.4m. What is its velocity at the bottommost point in its path?

Possible Answers:

vi.9m/s

48m/s

4.5m/s

17m/s

Explanation:

First solve for the potential free energy of the pendulum at the height of 2.4m.

PE = mgh

PE = (405kg)(10m/sⁱⁱ)(2.4m) = 9720J

This must be equal to the maximum kinetic energy of the object.

KE = ½mv²

9720J = ½mvⁱⁱ

Plug in the mass of the object (405 kg) and solve for 5.

9720J = ½(405kg)v²

v = 6.9m/due south

Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb upwards the hill using boots. Halfway up the fifty-meter colina, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the tiptop.

They then decide to sled downwards the hill, merely disagree about who will get first.

Scenario 1:

Sam goes down the hill first, claiming that he will attain a higher velocity. If Sally had gone first, Sam says they could collide.

Scenario 2:

Emerge goes down the hill first, challenge that she will experience lower friction and thus accomplish a college velocity. If Sam had gone starting time, Emerge says they could collide.

Scenario three:

Unable to agree, Sam and Sally tether themselves with a rope and get down together.

At the lesser of a neighboring hill, a neighbor watches Sally and Sam come down the hill. Sally is traveling 15m/s and Sam is traveling 10m/s. From the moment the neighbor begins watching, to just later on they both come up to a cease, who has prodigal more than heat in the form of friction? (Assume all friction is lost equally oestrus).

Possible Answers:

Sam, because he has more momentum

Sam, because he has greater kinetic energy

They dissipate equal amounts because the coeffecients of friction are the same for both

Sally, considering she has greater kinetic free energy

Sally, considering she has less momentum

Correct answer:

Sally, considering she has greater kinetic free energy

Explanation:

Sally has greater kinetic free energy in this instance than does Sam. From the moment when the neighbor begins watching nosotros can calculate the kinetic free energy. In one case stopped, all of the kinetic energy will have been dissipated.

Sally'south KE = one/2 (52kg) (15m/south)² = 5850J

Sam's KE = 1/2 (60kg) (10m/s)² = 3000J

All of this energy will be dissipated as friction before Sam and Sally come up to a stop.

Ignoring air resistance, which of the post-obit is true regarding the motion of a pendulum?

Possible Answers:

At the bottom of the oscillation potential energy is at a minimum and kinetic energy is at a minimum

At the bottom of the oscillation potential energy is at a minimum and kinetic energy is at a maximum

At the lesser of the oscillation potential free energy is at a maximum and kinetic free energy is at a maximum

At the bottom of the oscillation potential energy is at a maximum and kinetic free energy is at a minimum

Correct reply:

At the bottom of the oscillation potential free energy is at a minimum and kinetic energy is at a maximum

Explanation:

Pendulum

Energy must exist conserved through the move of a pendulum. Let point 1 represent the bottom of the oscillation and indicate 2 represent the top. At indicate 1, there is no potential energy, using point 1 as our "basis/reference," thus all of the organisation free energy is kinetic free energy. At indicate two, the velocity is zero; thus, the kinetic free energy is zero and all of the system energy is potential energy. At the highest point in the swing, potential energy is at a maximum, and at the lowest point in the swing, kinetic energy is at a maximum.

A 50kg bedrock drops from residuum off of a 100m cliff. Detect its velocity at before impact.

$\small g=10\frac{m}{s^2}$

Correct answer:

$43\frac{m}{s}$

Explanation:

Conservation of energy dictates that the initial free energy and final energy will be equal.

$PE_{i} + KE_{i} = PE_{f} + KE_{f}$

In this case, the boulder starts with zero kinetic free energy and ends with both kinetic and potential free energy.

$mgh_i+\frac{1}{2}mv_i^2=mgh_f+\frac{1}{2}mv_f^2$

$mgh_i+\frac{1}{2}m(0\frac{m}{2})^2=mgh_f+\frac{1}{2}mv_f^2$

$mgh_i=mgh_f+\frac{1}{2}mv_f^2$

We tin cancel the mass from each term and plug in the given values to solve for the velocity at a height of $\small 7m$ .

$(10\frac{m}{s^2})(100m)=(10\frac{m}{s^2})(7m)+\frac{1}{2}v_f^2$

$930\frac{m^2}{s^2}=\frac{1}{2}v_f^2$

$1860\frac{m^s}{s^2}=v_f^2$

$43\frac{m}{s}=v_f$

A block of wood is floating in space. A bullet is fired from a gun and hits the cake, embedding itself in the wood and generating estrus. Which of the following is conserved?

Possible Answers:

Mechanical energy

Kinetic energy

Momentum

Temperature

Caption:

Momentum is always conserved in a organization, when non experiencing external forces.

An inelastic collision tin can be identified if the two objects stick together afterwards the standoff occurs, such equally the bullet becoming embedded in the forest. In an inelastic standoff, kinetic energy is non conserved. Since mechanical energy is the sum of kinetic and potential energy, mechanical free energy is also not conserved. This lack of conservation is due to the conversion of some of the kinetic energy to heat and audio. The kinetic free energy decreases every bit the rut energy increases, resulting in a not-constant temperature.

A stone is dropped from a given pinnacle and allowed to hit the ground. The velocity of the rock is measured upon impact with the ground. Presume at that place is no air resistance.

In order for the velocity of the rock to be doubled before bear on, which of the post-obit is necessary?

Possible Answers:

Double the meridian from which the stone is dropped

Reduce past 75% the pinnacle from which the rock is dropped

Quadruple the top from which the rock is dropped

Reduce by 25% the top from which the rock is dropped

Halve the height from which the rock is dropped

Right answer:

Quadruple the height from which the rock is dropped

Explanation:

We can compare height and velocity by comparing the equations for potential and kinetic energy. This is possible because the rock initially has no kinetic energy (velocity is zero) and has no potential energy upon impact (top is naught). Using conservation of energy will yield the comparison below.

$mgh = \frac{1}{2}mv^{2}$

Considering velocity is squared in the equation for kinetic energy, it requires a quadrupling of height in lodge to double the velocity.

$mg(xh) = \frac{1}{2}m(2v)^{2}$

$mg(xh)=\frac{1}{2}m4(v)^2$

x=4

Caption:

iv. Choice one is correct because initially, all of the mechanical energy in the stone was potential energy and none was kinetic energy: ME = KE + PE. PE is "stored" in the rock-loma system by rolling the stone upwardly hill. It is obvious that it takes one-half equally much energy to whorl the rock half fashion up the hill, compared with rolling it to the tiptop. At the bottom of the hill, all of the PE volition have been converted into KE, given by the formula $\dpi{100} \small KE=\frac{1}{2}mv^{2}$

Since the PE was ½ mgh when rolling the stone half fashion up hill, it is the same as it rolls down hill.

An empty mining cart has a mass of 15kg and is traveling down a track that has a slope of $40^{\circ}$ to the horizontal. The cart is traveling at a rate of $10\frac{m}{s}$ when an operator notices a disturbance on the runway ahead and locks the wheels of the cart. What is the speed of the cart after it has traveled 20m with the wheels locked?

$\mu_s = 0.6,\ \mu_k=0.3$

$g=10\frac{m}{s^2}$

Right respond:

$16.3\frac{m}{s}$

Explanation:

Nosotros need the equation for conservation of energy for this problem:

E= U_i+K_i = U_f + K_f +F

Nosotros tin can eliminate final potential energy if nosotros fix the final tiptop to be zer. We are solving for final velocity, so let's rearrange for final kinetic energy.

K_f = U_i+K_i-F

Substituting our equations for each variable, we become:

$\frac{1}{2}mv_f^2 = mgh_i + \frac{1}{2}mv_i^2 - \mu_kNd$

Rearranging for terminal velocity we get:

$v_f = \sqrt{2gh_i+v_i^2-\frac{2\mu_kNd}{m}}$

If you derive this formula and are unsure of your work, simply check your units. Each term under the square root has units of $\frac{m^2}{s^2}$ , which volition ultimately give us units of $\frac{m}{s}$ , which is what nosotros want.

We have values for all variables except two: height and normal force.

Permit's calculate the height. We know that the between the initial and terminal states, the cart has traveled 20 meters down a slope of twoscore degrees. Therefore, nosotros can calculate height with the formula:

$sin(40^{\circ})=\frac{h_i}{20 m}$

h_i=(20m)sin(40^o)

h_i = 12.856 m

Now we just need to find the normal force. The following diagram will help visualize this calculation.

If you lot are unsure whether to apply sine or cosine, remember about it practically. As the angle gets less and less, the normal strength is going to go larger. This is feature of a cosine role.

Therefore, we can say that:

$N = mgcos(40^{\circ}) = (15kg)(10\frac{m}{s^2})cos(40^{\circ})= 114.9 N$

Now that we have all of our variables, it'south fourth dimension to plug and chug:

$v_f=\sqrt{(2\cdot10\cdot12.856)+(10)^2-\frac{2\cdot0.3\cdot114.9\cdot20}{15}}$

$v_f=\sqrt{257.1+100-91.9} = 16.3 \frac{m}{s}$

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Review Questions on Jeat Transfer and Conservation Energy

All MCAT Physical Resources

All MCAT Physical Resource

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