Ap Physics 1 Pulley Problems

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Consider the following system:

Frictionless pully

Two masses, A and B, are attached to the end of a rope that runs through a frictionless pulley.

The two masses are initially held at residuum. When they are released, what is the initial instantaneous dispatch of mass A, and in what direction? Neglect air resistance and any frictional forces.

m_A = 20kg, m_B = 12kg

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

Right answer:

$2.5\frac{m}{s^2}$

Explanation:

We can utilise Newton's second police force:

$F_{net}=ma$

There are two forces interim on the arrangement: gravity on each mass. For simplicity sake, we will examine these forces with respect to mass. Therefore, the strength on mass B becomes an upwards tension strength on mass A. At present we demand to clarify which direction is positive. For this problem, we'll say that a downward force is positive, and an upwards force is negative.

Nosotros'll outset with the gravitational force:

F_g = m_Ag

Now for tension. The tension force is just the gravitational strength practical to mass B:

F_T = -m_Bg

Note that information technology's negative because information technology is in the upwardly direction

Now adding these together to go the net force:

$F_{net}= m_Ag-m_Bg = g(m_A-m_B)$

Substituting this back into the original equation, we get:

ma = g(m_A-m_B)

Where m is the combination of both masses:

(m_A+m_B)a=(m_A-m_B)g

Rearrange for dispatch:

$a = \frac{g(m_A -m_B)}{m_A+m_B}$

We know each value, and so nosotros can solve the problem

$a = \frac{\left ( 10\frac{m}{s^2}\right )(20kg-12kg)}{20kg+12kg}$

$a = 2.5\frac{m}{s^2}$

Consider the following system:

Frictionless pully

2 masses, A and B, are attached to the stop of a rope that runs through a frictionless pulley.

The two masses are initially at rest. The pulley has a radius r = 0.25m . The masses are released. After t = 3s , what is the rotational velocity of the caster? Fail air resistance and any frictional forces.

m_A = 5kg, m_B = 2kg

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

Right answer:

$72\frac{1}{s}$

Explanation:

In lodge to determine rotational velocity of the pulley, we demand the linear velocity of the outside of the pulley, which is simply the linear velocity of the rope/masses. We can determine that using the dispatch of the masses and the length of fourth dimension given. Therefore, nosotros will brainstorm with Newton's 2nd police practical to one of the two masses. We volition arbitrarily utilize information technology to mass A:

$F_{net}=m_Aa$

There are two forces applied to mass A, gravity and tension from the rope. We will assume that a downward force is positive, and an upwardly force is negative. First, let's kickoff with gravity:

F_g = m_Ag

Next nosotros have tension, which is but the gravitational force applied on mass B:

F_T = -m_Bg

Adding these together to get net force:

$F_{net}= m_Ag-m_Bg=g(m_A-m_B)$

Substituting this into the original equation:

m_Aa = g(m_A-m_B)

Rearranging for acceleration:

$a = \frac{g(m_A-m_B)}{m_A}$

At present that nosotros have acceleration, nosotros tin use it to decide linear velocity after the stated length of time:

$v_f = at = \frac{gt(m_A-m_B)}{m_A}$

Now we can employ this to determine rotational velocity:

$w_f = \frac{gt(m_A-m_B)}{rm_A}$

We know all of these values, so time to plug and chug:

$w_f = \frac{\left ( 10\frac{m}{s^2}\right )(3s)(5kg-2kg)}{(0.25m)(5kg)}$

$w_f = 72\frac{1}{s}$

Consider the post-obit organisation:

Frictionless pully

Two masses, A and B, are fastened to the end of a rope that runs through a caster.

The system is at remainder. The caster has become quit old and has develop a thick coat of rust. If m_A = 4kg , what range of mass B will result in the organisation remaining at rest if the static frictional force within the pulley is 15N ?

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

Correct answer:

$2.5kg\leq m_B \leq 5.5kg$

Explanation:

We will arroyo this problem by looking at the pulley. At that place are 3 forces applied to the pulley, tension from each weight every bit well as its internal friction strength. To clarify, we will denote that forces in the counterclockwise direction will be positive and forces in the clockwise direction are negative. Also, we need to recollect that the internal static friction tin can be applied in both the clockwise and counterclockwise management. This will create 2 scenarios. Allow's kickoff began when friction is applied in the counter clockwise direction. We will offset with Newton'due south 2nd law:

$F_{net}=ma$

Since nosotros are asked to detect when the arrangement remains at rest, that means the net force on the organization is 0:

$F_{net}=0$

At present we can begin looking at the iii forces. Get-go, we have tension created from mass A. This volition be positive since it is in the counterclockwise direction relative to the caster:

F_A=m_Ag

Next we have mass B which will be negative since it'due south in the opposite management:

F_B = -m_Bg

Lastly we have static friction which is positive in this scenario:

F_F=15N

Now adding these together to get equation (1):

$F_{net} = 15N +m_Ag-m_Bg= 0$

Rearranging for mass B:

$m_B = \frac{15N+m_Ag}{g}$

$m_B = \frac{15N + (4kg)\left (10\frac{m}{s^2}\right )}{10\frac{m}{s^2}}$

m_B = 5.5kg

Note that anything greater will issue in the system moving.

Now moving onto the 2nd scenario. Permit's just jump back to equation (ane) and simply reverse the management of static friction:

$F_{net}=\frac{-15N+m_Ag}{g}$

$m_B = \frac{-15N+(4kg)\left ( 10\frac{m}{s^2}\right )}{10\frac{m}{s^2}}$

m_B = 2.5kg

Notation that anything less than this will result in the arrangement moving. Therefore, the range of mass B that will go along the system stationary is:

$2.5kg\leq m_B \leq 5.5kg$

In a match of tug of state of war, the Red Team can pull in the positive x-direction with a combined force of $1250\textup{ N}$ . Pulling against them in exactly the opposite management is the Blue Squad with a combined forcefulness of $950\textup{ N}$ . What is the internet forcefulness on the rope?

Right answer:

$300\textup{ N}$

Explanation:

We will use the sum of forces to solve this equation.

$F_{net}=F_{Red}+F_{Blue}$

$F_{Net}=1250N+(-950N)$

$F_{Net}=300N$

The Scarlet Team volition win this match of tug of state of war past pulling with 300N of actress force compared to the Blue Squad.

What is the net force required to accelerate a race car that weighs $3009\textup{ lbs}$ , that can go from rest to $62\ \frac{\textup{mi}}{hr}$ in $3.70\textup{ s}$ ?

Correct answer:

$1.02*10^{4}\textup{ N}$

Explanation:

To solve this question use Newton's second constabulary and the sum of forces to solve. First, before we solve we have to become our given values into the proper units.

$62\frac{mi}{hr}=\frac{5280ft}{1mi}=\frac{1m}{3.28ft}=\frac{1hr}{60min}=\frac{1min}{60s}=27.72\frac{m}{s}$ $\frac{1min}{60s}=27.72\frac{m}{s}$

$3009lbs=\frac{0.454kg}{1lbs}=1366.1kg$

At present we have to solve for the sport automobile's dispatch

$a=\frac{\Delta V}{\Delta t}$

$a=\frac{27.72\frac{m}{s}-0\frac{m}{s}}{3.70s-0s}$

$a=7.49\frac{m}{s^{2}}$

Now use the cyberspace forces to solve.

$F_{Net}=ma$

$F_{Net}=1366.1Kg(7.49\frac{m}{s^{2}})$

$F_{Net}=1.02*10^{4}N$

In an experiment on the Moon, the acceleration due to gravity was adamant to be $\frac{1}{6}$ the normal value at sea level on Earth. What is the weight of a baseball game on the Moon that weighs $1.460\textup{ N}$ at sea level on Earth?

Correct answer:

$0.243\textup{ N}$

Explanation:

w=ma

Looking at the equation for weight. Notice, weight and acceleration are directly proportional. Thus what happens to the acceleration also happens to the weight.

$w_{earth}=1.460N=ma$

$1.460N=m(9.8\frac{m}{s^{2}})$

m=0.149Kg

The acceleration due to gravity on the Moon is $\frac{1}{6}$ that on Earth.

$g_{earth}=9.8\frac{m}{s^{s}}=g_{moon}=9.8\frac{m}{s^{2}}(\frac{1}{6})=1.63\frac{m}{s^{2}}$

$w_{moon}=0.149Kg(1.63\frac{m}{s^{2}})$

$w_{moon}=0.243N$

The ball will weigh $\frac{1}{6}$ information technology's original value on Globe.

If a toy auto with a mass of $500\textup{ g}$ has an engine that applies a constant forcefulness of $5\textup{ N}$ . How long will it accept the toy automobile to reach its max velocity of $20\ \frac{\textup{m}}{\textup{s}}$ when released from rest? Assume the car is on a frictionless surface and you may also neglect air resistance.

Correct answer:

$2.0\textup{ s}$

Caption:

Apply Newton'south second law to determine the automobile'south dispatch.

F=ma

5N=0.5Kg(a)

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

At present use a kinematic equation to solve for the time required.

$V_{f}=V_{i}+a(\Delta t )$

$20\frac{m}{s}=0\frac{m}{s}+10\frac{m}{s^{2}}(t_{f}-0s)$

$t_{f}=2.0seconds$

Tony applies a constant horizontal force of $140\textup{ N}$ on a bag that weighs $245\textup{ N}$ and is resting on a flat surface, experiencing a coefficient of kinetic friction of 0.35 . What is the net forcefulness acting on the handbag?

Correct answer:

$54.2\textup{ N}$

Explanation:

The bag is on a flat surface, thus the normal force is equal to the weight of the bag.

$w=245N=F_{n}$

Use the equation for kinetic friction to solve for the kinetic friction.

$F_{k}=\mu _{s}(F_{n})$

$F_{k}=0.35(245N)$

$F_{k}=85.8N$

Now use the sum of forces to solve for the net force acting on the handbag.

$F_{net}=F_{pull}+(F_{k})$

$F_{net}=140N+(-85.8N)$

$F_{net}=54.2N$

There will exist a total of 54.2N acting on the bag as it is pulled across the ground.

A $205\textup{ lb}$ baseball actor slides, making contact with the ground causing a forcefulness of kinetic friction measured at $250\textup{ N}$ . What is the acceleration that acts on this superstar due to friction during the slide?

$1\textup{ lb}=0.453\textup{ kg}$

Correct answer:

$-2.7\ \frac{\textup{m}}{\textup{s}^{2}}$

Caption:

Firstly, put the given values into the proper units.

$205lbs=\frac{0.453Kg}{1lbs}=92.87Kg$

Now utilise Newton's second law to determine the acceleration.

F=ma

250N=92.87Kg(a)

$a=2.69\frac{m}{s^{2}}$

He will decelerate at $2.7\frac{m}{s^{2}}$ .

Kaden lives on the fourth floor at an flat complex. Instead of making multiple trips upwards and downwards the stairs he elects to apply a rope and bucket organisation to get his groceries up to his balcony. If Kaden applies $98\textup{ N}$ of tension on the rope tied to the bucket. What mass (in kilograms) of groceries tin he carry in each load? Assume that both the bucket and rope are weightless and neglect friction.

Correct answer:

$10\textup{ kg}$

Explanation:

The bucket is not accelerating in either direction so the tension strength must exist equal to the weight of the groceries.

T=W

Multiplying mass and gravity gives weight

T=mg

Plug in and solve for the mass

$98N=m(9.8\frac{m}{s^{2}})$

m=10Kg

Kaden can acquit a maximum of 10Kg of groceries in each load upward to his balcony.

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