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Nadusha1986 [10]
3 years ago
7

A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,

place a block of mass m on top of the compressed spring, and then release the block. The spring launches the block upward, and the block rises to a maximum height some distance above the now-relaxed spring. Find the speed of the block just as it loses contact with the spring.
Physics
1 answer:
shusha [124]3 years ago
6 0

Answer:

v=d\sqrt{\frac{k}{m}}

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

U_{0}+K_{0}=U_{f}+K_{f}

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

U_{0}=K_{f}

The initial potential energy of the spring is given by the equation:

U_{0}=\frac{1}{2}kd^{2}

the Kinetic energy of the block is then given by the equation:

K_{f}=\frac{1}{2}mv_{f}^{2}

so we can now set them both equal to each other, so we get:

=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

kd^{2}=mv_{f}^{2}

so now we can solve this for the final velocity, so we get:

v=d\sqrt{\frac{k}{m}}

You might be interested in
Energy conservation
balandron [24]

Answer:

7.328m/s

Explanation:

Given parameters:

height of table = 0.68m

final velocity of the ball = 6m/s

Unknown:

Initial velocity of ball = ?

Solution:

To solve this problem, we are going to employ the appropriate motion equation.

We must understand that this fall occurs in the presence of gravity;

            V = U + 2gH

Where;

 V is the final velocity

 U is the initial velocity

 g is the acceleration due to gravity

 H is the height of the pool table

    Since U is the unknown, let us make it the subject of the expression;

           U = V - 2gH

        U = 6 - (2 x 9.8 x 0.68)  = 7.328m/s(deceleration)

7 0
3 years ago
Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh
Alinara [238K]

Answer:

a) k_{avg}=6.22\times 10^{-21}

b) k_{avg}=8.61\times 10^{-21}

c)  k_{mol}=3.74\times 10^{3}J/mol

d)   k_{mol}=5.1\times 10^{3}J/mol

Explanation:

Average translation kinetic energy (k_{avg}) is given as

k_{avg}=\frac{3}{2}\times kT    ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8  

k_{avg}=6.22\times 10^{-21}J

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416  

k_{avg}=8.61\times 10^{-21}J

c ) The translational kinetic energy per mole of an ideal gas is given as:

       k_{mol}=A_{v}\times k_{avg}

here   A_{v} = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

          k_{mol}=3.74\times 10^{3}J/mol

d) now at T = 143° C

        k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}

          k_{mol}=5.1\times 10^{3}J/mol

8 0
3 years ago
Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water
tangare [24]

Answer:

T_b=107.3784\ ^{\circ}C

Explanation:

Given:

  • thickness of the base of the kettle, dx=0.52\ cm=5.2\times 10^{-3}\ m
  • radius of the base of the kettle, r=0.12\ m
  • temperature of the top surface of the kettle base, T_t=100^{\circ}C
  • rate of heat transfer through the kettle to boil water, \dot Q=0.409\ kg.min^{-1}
  • We have the latent heat vaporization of water, L=2260\times 10^3\ J.kg^{-1}
  • and thermal conductivity of aluminium, k=240\ W.m^{-1}.K^{-1}

<u>So, the heat rate:</u>

\dot Q=\frac{0.409\times 2260000}{60}

\dot Q=15405.67\ W

<u>From the Fourier's law of conduction we have:</u>

\dot Q=k.A.\frac{dT}{dx}

\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}

where:

A= area of the surface through which conduction occurs

T_b= temperature of the bottom surface

15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}

T_b=107.3784\ ^{\circ}C is the temperature of the bottom of the base surface of the kettle.

6 0
3 years ago
"Two uniform identical solid spherical balls each of mass M and radius R" and moment of inertia about its center 2/5 MR2 are rel
adelina 88 [10]

Answer:

he sphere that uses less time is sphere A

Explanation:

Let's start with ball A, for this let's use the kinematics relations

        v² = v₀² - 2g (y-y₀)

indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0

         v² = 0 - 2 g (0- y₀)

         v = \sqrt{2g y_o}

         v = \sqrt{2 \ 9.8\ H}

         v = 4.427 √H

Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy

starting point. At the highest point, before you start to move

         Em₀ = U = m g y

final point. At the bottom of the ramp

         Em_f = K = ½ m v² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

          Em₀ = Em_f

          mg H = ½ m v² + ½ I w²

angular and linear velocity are related

          v = w r

          w = v / r

the momentorot of inertia indicates that it is worth

          I = \frac{2}{5} m r²

we substitute

           m g H = ½ m v² + ½ (\frac{2}{5}  m r²) (\frac{v}{r} )²

           gH = \frac{1}{2}  v² + \frac{1}{5}  v² = \frac{7}{10}  v²

           v = \sqrt{\frac{10}{7} \ g H}

           v = \sqrt{ \frac{10}{7}  \ 9.8 \ H}

           v=3.742 √H

Taking the final speeds of the sphere, let's analyze the distance traveled, sphere A falls into the air, so the distance traveled is H.  The ball B rolls in a plane, so the distance (L) traveled can be found with trigonometry

           sin θ = H / L

           L = H /sin θ

we can see that L> H

In summary, ball A arrives with more speed and travels a shorter distance, therefore it must use a shorter time

Consequently the sphere that uses less time is sphere A

5 0
3 years ago
Write a difference between force and work? ​
Nesterboy [21]

Answer:

force is a derived quantity

work is a fundamental quantity

2.Force is measured in Newton why work is measured in Newton per seconds (N/s)

Explanation:

no

3 0
3 years ago
Read 2 more answers
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