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
aleksley [76]
3 years ago
5

A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the

bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest?
Physics
1 answer:
Aleks04 [339]3 years ago
7 0

Answer:

The block slides on the horizontal surface 25 m before coming to rest.

Explanation:

Hi there!

For this problem, we have to use the energy-conservation theorem. Initially, the block has only gravitational potential energy (PE) that can be calculated as follows:

PE = m · g · h

Where:

m = mass of the block.

g = acceleration due to gravity.

h = height at which the block is located.

As the block starts to slide down the track, its height diminishes as well as its potential energy. Due to the conservation of energy, energy can´t disappear, so the loss of potential energy is compensated by an increase of kinetic energy (KE). In other words, as the block slides, the potential energy is converted into kinetic energy. The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

m = mass of the block.

v = speed of the block.

Then, at the bottom of the ramp, the kinetic energy of the block will be equal to the potential energy that the block had at the top of the ramp.

Initial PE = KE at the bottom

When the block starts sliding horizontally, friction force does work to stop the block. According to the energy-work theorem, the change in the kinetic energy of an object is equal to the net work done on that object. In other words, the amount of work needed to stop the block is equal to its kinetic energy. Then, the work done by friction will be equal to the kinetic energy of the block at the bottom, that is equal to the potential energy of the block at the top of the track:

initial PE = KE at the bottom = work done by friction

The work done by friction is calculated as follows:

W = Fr · Δx

Where:

W = work

Fr = friction force.

Δx = traveled distance.

And the friction force is calculated as follows:

Fr = μ · N

Where:

μ = coefficient of friction.

N = normal force.

Since the block is not accelerated in the vertical direction, in this case, the normal force is equal to the weight (w) of the block:

Sum of vertical forces = ∑Fy = N - w = 0 ⇒N = w

And the weight is calculated as follows:

w = m · g

Where m is the mass of the block and g the acceleration due to gravity.

Then, the work done by friction can be expressed as follows:

W = μ · m · g · Δx

Using the equation:

intial PE = work done by friction

m · g · h = μ · m · g · Δx

Solving for Δx

h/μ = Δx

5.0 m / 0.20 = Δx

Δx = 25 m

The block slides on the horizontal surface 25 m before coming to rest.

You might be interested in
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin
lyudmila [28]
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
W= \int\limits^{0.540m}_{0} {F} \, dx =  \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =
=16000x+10000  \frac{x^2}{2} - 26000  \frac{x^3}{3}
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
W=16000(0.540m)+10000  \frac{(0.540m)^2}{2} - 26000  \frac{(0.540m)^3}{3}  =
=8733 J=8.73 kJ

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
=12280 J=12.28 kJ
5 0
3 years ago
Un satélite geoestacionario se encuentra a una distancia de 120.000 km sobre la superficie de Júpiter. Determine: a. El periodo
Lisa [10]

Answer:

a) a geostationary satellite is that it is always at the same point with respect to the planet,

b) f = 2.7777 10⁻⁵ Hz

c)                           d)   w = 1.745 10⁻⁴ rad / s

Explanation:

a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet

  •                T = 10 h (3600 s / 1h) = 3.6 104 s

b) the period the frequency are related

                T = 1 / f

                 f = 1 / T

                 f = 1 / 3.6 104

                 f = 2.7777 10⁻⁵ Hz

c) the distance traveled by the satellite in 1 day

The distance traveled is equal to the length of the circumference

                 d = 2pi (R + r)

                 d = 2pi (69 911 103 + 120 106)

                 d = 1193.24 m

d) the angular velocity is the angle traveled between the time used.

                 .w = 2pi /t

                  w = 2pi / 3.6 10⁴

                  w = 1.745 10⁻⁴ rad / s

how fast is

                  v = w r

                  v = 1.75 10-4 (69.911 106 + 120 106)

                  v = 190017 m / s

5 0
3 years ago
A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th
ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

<u>ac = 3.92 m/s²</u>

5 0
3 years ago
A sailfish swims 120 km/hr. How far will it travel in 8.0 minutes?
Rasek [7]

Answer:

16km

Explanation:

First change the minutes into hours then multiply by the distance.

(8÷60)×120=16km

5 0
4 years ago
Calculate the acceleration of a car if it's velocity increases from 15 m/s to 75m/s in 5 seconds. ​
Andreas93 [3]
  • initial velocity=u=15m/s
  • Final velocity=v=75m/s
  • Time=t=5s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{75-15}{5}

\\ \sf\longmapsto Acceleration=\dfrac{60}{5}

\\ \sf\longmapsto Acceleration=12m/s^2

7 0
3 years ago
Other questions:
  • Mary and her younger brother Alex decide to ride the 17 -foot-diameter carousel at the State Fair. Mary sits on one of the horse
    6·1 answer
  • What is an example of chemical potential energy in humans?
    15·1 answer
  • 1. Calculate the force of gravity in newtons if your weight is 110 lbs?
    5·2 answers
  • A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules (with the poten
    12·1 answer
  • You must exert a force of 4.5 N on a book to slide it across a table. If you do 2.7J
    10·1 answer
  • Heat naturally flows from an object that has a _______________ temperature to an object that has a _______________ temperature.
    11·2 answers
  • A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied
    11·1 answer
  • We know that there is a relationship between work and mechanical energy change. Whenever work is done upon an object by an exter
    8·1 answer
  • HELLO I NEED YOU HELP WITH THIS SCIENCE QUESTION NO LINKS!!!
    5·1 answer
  • There are two factors that limit how much he can bake in a week: He only wants to work for 40 hours a week and he only has one o
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!