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
AfilCa [17]
3 years ago
14

The 3.0 kg object below is released from rest at a height of 5.0 m on a curved frictionless ramp. At the foot of the ramp is a s

pring of force constant k = 500 N/m. The object slides down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. a) Find x. b) what happens to the object after it comes to rest?
Physics
1 answer:
scoray [572]3 years ago
7 0

Answer:

a) The compressing distance is 0.8 m.

b) After the object comes to rest, it will be released with an initial speed of 7.1 m/s and will reach a height of 5.0 m.

Explanation:

Hi there!

a)Initially, the object has only gravitational potential energy that is calculated as follows:

PE = m· g · h

Where:

PE = potential energy.

m = mass.

g = acceleration due to gravity.

h = height.

Then:

PE = 3.0 kg · 9.8 m/s² · 5.0 m = 1.5 × 10² J

When the object reaches the spring and compresses it, all that potential energy is converted into elastic potential energy (there is no energy dissipation as heat because there is no friction). The elastic potential energy is calculated as follows:

EPE = 1/2 · k · x²

Then, the elastic potential energy is equal to the initial gravitational energy:

PE = EPE = 1.5 × 10² J = 1/2 · 500 N/m · x²

Solving for x:

1.5 × 10² Nm / (1/2 · 500 N /m) = x²

0.6 m² = x²

x = 0.8 m

The compressing distance is 0.8 m.

b) After the object comes to rest, the object will be released with a kinetic energy equal to the elastic potential energy. We can calculate the initial velocity of the object after it is released with the following equation:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass.

v = speed.

Since initially EPE = KE:

1.5 × 10² J = 3.0 kg · v²

√( 1.5 × 10² J / 3.0 kg) = v

v = 7.1 m/s

Then, after the object comes to rest, it will be released with an initial speed of 7.1 m/s and will reach a height of 5.0 m.

You might be interested in
The viscosities of several liquids are being compared. All the liquids are poured down a slope with equal path lengths. The liqu
Vladimir [108]

Answer:

Move slowly and reach bottom later.

Explanation:

Viscosity is termed as the thickness or consistency of any liquid or semi liquid. It is related to the internal friction of the substance.

When several liquids are poured down with equal path lengths then the liquid will high viscosity will reach the bottom latter while one with less viscosity.

The internal friction of the molecules tends to keep them together making its consistency more thick. Thus when it will slope down from a certain height it will take more time to reach down.

5 0
3 years ago
A 70 kg base runner begins his slide into second base while moving at a speed of 4.35 m/s. He slides so that his speed is zero j
sladkih [1.3K]

Answer

given,

Mass of the runner, M = 70 Kg

speed of the runner on the second base = 4.35 m/s

speed at the base = 0 m/s

Acceleration due to gravity,g = 9.8 m/s²

a) magnitude of mechanical energy lost

  Mechanical energy lost is equal top gain in kinetic energy

   ME_{Lost}=\dfrac{1}{2}mv^2

   ME_{Lost}=\dfrac{1}{2}\times 70\times 4.35^2

   ME_{Lost}=662.29\ J

b) Work done = Force x displacement

    W = F. x

     F = μ mg

    W = μ mg . x

Work done is equal to 662.29 J

  x=\dfrac{W}{\mu m g}

using the coefficient of the friction,μ = 0.7

  x=\dfrac{662.29}{ 0.7\times 70\times 9.8}

     x = 1.38 m

Hence, the runner will slide to 1.38 m.

3 0
3 years ago
Sound travels at a speed of 1188 km in one hour. How many meters will it travel in one second?
Nana76 [90]
The answer to this question would be 330 m/sec
7 0
3 years ago
Read 2 more answers
4.
zimovet [89]

Answer:

hindi ko gets

Explanation:

ang gulo naman nyan

8 0
2 years ago
An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm
stepan [7]

The object rises to a height of 20.4 m. So option C is correct.

Explanation:

initial velocity= Vi=20 m/s

final velocity at the top= Vf=0

acceleration= g=-9.8 m/s²

Vf²=Vi²+2gh

0= (20)²+2 (-9.8)h

-200=-9.8h

h=200/9.8

h=20.4 m

Thus the object rises to a height of 20.4 m

7 0
3 years ago
Other questions:
  • A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
    7·1 answer
  • A magic school bus travelled along this path. What is the magnitude of the total
    13·1 answer
  • Which best describe insane
    5·2 answers
  • An information signal consists of a 25 Hz and a 75 Hz sine waves summed together. It is sampled at a frequency of 500 Hz, the hi
    11·1 answer
  • On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 2.50 times as far as he wo
    13·1 answer
  • Michael has a substance that he puts in Container 1. The substance has a volume of 5 cubic meters. He then puts the substance in
    13·1 answer
  • how do u calculate the kinetic energy of a ball of mass 0.25kg being kicked vertically upwards with a speed of 5m/s​
    14·1 answer
  • Which of the following statements is
    15·2 answers
  • How is it that even light precipitation can still cause the collection of large amounts of water.
    12·1 answer
  • Johanna is studying what happens to the energy as a ball rolls down a ramp. What is one form of energy that she is studying? gra
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!