1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mila [183]
3 years ago
14

Block 1, with mass m1 and speed 3.6 m/s, slides along an x axis on a frictionless floor and then undergoes a one-dimensional ela

stic collision with stationary block 2, with mass m2 = 0.40m1. The two blocks then slide into a region with a coefficient of kinetic friction of 0.50 where they stop. How far into that region do the two blocks slide?(a) block 1 m(b) block 2 m
Physics
1 answer:
irina1246 [14]3 years ago
3 0

Answer:

a) The block 1 slides 0.24 m into the rough region.

b) The block 2 slides 2.7 m

Explanation:

Hi there!

First, let´s find the final velocity of each block. With that velocities, we can calculate the kinetic energy of each block. The kinetic energy of the blocks will be equal to the work done by friction to stop them. From the equation of work, we can calculate the distance traveled by the blocks.

Since the collision is elastic, the momentum and kinetic energy of the system composed of the two blocks is constant.

The momentum of the system is calculated as the sum of the momenta of each block:

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

Where:

m1 and m2 = mass of blocks 1 and 2 respectively.

v1 and v2 = velocity of blocks 1 and 2 respectively.

v1´ and v2´ = final velocity of blocks 1 and 2 respectively.

Using the data we have, we can solve the eqaution for v1´:

m1 · 3.6 m/s + 0.40 m1 · 0 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s · m1 = m1 · v1´ + 0.40 m1 · v2´

3.6 m/s = v1´ + 0.40 v2´

v1´ = 3.6 m/s - 0.40 v2´

The kinetic energy of the system also remains constant:

1/2 m1 · (v1)² + 1/2 m2 · (v2)² = 1/2 m1 · (v1´)² + 1/2 m2 · (v2´)²

Multiply by 2 both sides of the equation:

m1 · (v1)² + m2 · (v2)² = m1 · (v1´)² + m2 · (v2´)²

Let´s replace with the data:

m1 · (3.6 m/s)² + 0.40 m1 · 0 = m1 · (v1´)² + 0.40 m1 (v2´)²

divide by m1:

(3.6 m/s)² = (v1´)² + 0.40 (v2´)²

Replace v1´ = 3.6 m/s - 0.40 v2´

(3.6 m/s)² = (3.6 m/s - 0.40 v2´)² + 0.40 (v2´)²

Let´s solve for v2´:

(3.6 m/s)² = (3.6 m/s)² - 2.88 v2´ + 0.16 (v2´)² + 0.40 (v2´)²

0 = 0.56 (v2´)² - 2.88 v2´

0 = v2´(0.56 v2´ - 2.88)   v2´ = 0 (the initial velocity)

0 = 0.56 v2´ - 2.88

2.88/0.56 = v2´

v2´ = 5.1 m/s

Now let´s calculate v1´:

v1´ = 3.6 m/s - 0.40 v2´

v1´ = 3.6 m/s - 0.40 (5.1 m/s)

v1´ = 1.56 m/s

Now, let´s calculate the final kinetic energy (KE) of each block:

a) Block 1:

KE = 1/2 · m1 · (1.56 m/s)² = m1 · 1.2 m²/s²

The work done by friction is calculated as follows:

W = Fr · s

Where:

Fr = friction force.

s = traveled distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction.

And the normal force is calculated in this case as:

N = m1 · g

Where g is the acceleration due to gravity.

Then, the work done by friction will be:

W = m1 · g · μ · s

The kinetic energy of an object is the negative work that must be done on that object to bring it to stop. Then:

m1 · 1.2 m²/s² = m1 · g · μ · s

Solving for s:

s = m1 · 1.2 m²/s²  / m1 · g · μ

s = 1.2 m²/s²/ 9.8 m/s² · 0.50

s = 0.24 m

The block 1 slides 0.24 m into the rough region.

b) For block 2 the kinetic energy will be the following:

KE = 1/2 · 0.4 · m1 · (5.1 m/s)² = m1 · 5.2 m²/s²

The friction force will be:

Fr = 0.4 m1 · g · μ

And the work done will be:

W = 0.4 m1 · g · μ · s

Since W = ΔKE,

Then:

m1 · 5.2 m²/s² = 0.4 m1 · g · μ · s

Solving for s:

5.2 m²/s²/(0.4 · g · μ) = s

s =  5.2 m²/s²/(0.4 · 9.8 m/s² · 0.50)

s = 2.7 m

The block 2 slides 2.7 m

You might be interested in
Differences between concave and convex meniscus​
BARSIC [14]

Explanation:

A concave meniscus,(normally seen) occurs when the molecules of the liquid are attracted to those of the container. This occurs with water and a glass tube. A convex meniscus occurs when the molecules have a stronger attraction to each other than to the container, as with mercury and glass.

4 0
3 years ago
Read 2 more answers
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
3 years ago
un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l
Dmitriy789 [7]

Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

Distance cover in 13 seconds = 487.56 meter

8 0
2 years ago
Which of the following is NOT an argument for showing that the Earth must be round: a. during an eclipse of the Moon, the shadow
Hitman42 [59]

Answer: The correct option is option E (the Sun is seen blocking different constellations in the course of a year.

Explanation:

The earth, which is one of the planets of the solar system that supports life, is shperical in shape. The spherical ( round) shape of the earth is marked by the intervening highlands and oceans on its surface.

Evidence to show that the earth is shperical are:

--> The Lunar eclipse: During an eclipse of the Moon, the shadow of the Earth is always seen to be round.

--> Ships Visibility: When ships travel a large distance away, we see their hulls disappear first and their masts disappear last.

-->Altitude of Polaris (North Star): The height of the North Star changes as we travel to different latitudes. That is ,increases as you move toward the North pole, or decreases as you move toward the equator.

--> Aerial photographs: Photographs of the Earth from space always show a round body.

The statement that doesn't prove that the earth is spherical in shape is (the Sun is seen blocking different constellations in the course of a year). The sun is seen in front of stars blocking different constellation in a year because the earth orbits round the sun in a year and not that it is shperical in shape.

4 0
2 years ago
½H2(g) + ½I2(g) → HI(g), ΔH = + 26 kJ/mole 117 kJ/mole + C(s) + 2S(s) → CS2(g) The temperature of the surroundings will:
Alona [7]
From the above reaction the temperature of the surroundings will increase.
8 0
3 years ago
Read 2 more answers
Other questions:
  • A brick is released with no initial speed from the roof of a building and strikes the ground in 1.80 s , encountering no appreci
    10·1 answer
  • In photography what are clouds best described as?
    15·1 answer
  • When two or more elements chemically combine the result is?
    14·1 answer
  • Match each characteristic with the type of wave it corresponds to. Longitudinal waves Transverse waves The direction the wave tr
    9·2 answers
  • Explain why the cyclist must be fed in order to continue to pedal? answer fast please​
    7·1 answer
  • The force it would take to accelerate a 900-kg car at a rate of 3 m/s2 is
    10·2 answers
  • This is the equation for the formation of magnesium chloride. Mg(s) + 2HCl(l) → MgCl2(aq) + H2(g) Which are the reactants and th
    13·1 answer
  • A student drops an object from rest above a force plate that records information about the force exerted on the object as a func
    8·2 answers
  • A bowling ball hits two standing bowling pins at the same time. Which of the following is true? Assume that all collisions are e
    5·1 answer
  • Nancy ran a distance of 5km in 30 minutes. What is her speed in meters(m) per hour)h)
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!