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-Dominant- [34]
3 years ago
5

A light spring with a force coefficient 11.85 N/m is compressed by 14 cm as it is held between a 0.27 kg block on the left and a

0.6 kg block on the right, both resting on a horizontal surface. The spring exerts a force on each of the blocks, tending to push them apart. The blocks are then released from rest.Find the acceleration of each block if the coefficient of friction between the blocks and the surface is:
[Note: This question may require you to perform your calculations to a few extra digits of precision to get the answer to within 1%.]

(a) μ = 0: Left______ Right _______

(b) μ = 0.158: Left______ Right _______
Physics
1 answer:
lilavasa [31]3 years ago
4 0

Answer:

A) Left = 6.14 m/s2 Right=2.765 m/s2   B) Left = 4.59 m/s2 Right= 1.215 m/s2

Explanation:

In the question we are given the spring constant which is 11.85 N/m and the compression of the spring is 14 cm. There are two blocks in front of the springs which are 0.27 kg and 0.6 kg respectively.

First, we need to calculate how much force the spring are going to exert on the masses when they are released. Since this force is not going to change for both cases, we only need to do it once.

The formula for calculation the force is F = k.x where k is the spring constant or force coefficient 11.85 N/m and the x is the compression which is 14 cm or 0.14 meters. If we put them in the equation we can find that F = 1.659 N

A) In the first case scenario, where the friction is equal to 0, we can use the formula F=m.a where F is the force applied by the release of the spring, m is the mass of the block and a is the acceleration.

For the first block, when we put 1.659 N and 0.27kg in the equation, a is calculated to be 6.14 m/s2.

For the second block, the same force of 1.659 N and 0.6 kg, a is calculated to be 2.765 m/s2.

B) In the second case scenario, where the friction is equal to 0.158, we first need to calculate its effect on each block. We need to use the formula      Fk = μ.N where μ is the friction constant and N is the normal force of the block which is m.g (where g = 9,81 m/s2).

The friction force for the first block is calculated to be 0.4184 N and the friction force for the second block is calculated to be 0.9299 N.

The total force for the first block is 1.659 N - 0.4184 N = 1.2405 N.

The total force for the second block is 1.659 N - 0.9299 N = 0.7290 N.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the first block as 1.2405 = 0.27 x a  and a = 4.59 m/s2.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the second block as 0.7290 = 0.6 x a  and a = 1.215 m/s2.

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The final velocity of the bullet+block is 0.799 m/s

Explanation:

We can solve this problem by applying the principle of conservation of momentum: in fact, the total momentum of the bullet-block system must be conserved before and after the collision.

Mathematically, we can write:

mu+MU=(m+M)v

where

m = 0.001 kg is the mass of the bullet

u = 800 m/s is the initial velocity of the bullet

M = 1 kg is the mass of the block

U = 0 is the initial velocity of the block (initially at rest)

v is the final combined velocity of the bullet and the block

Solving the equation for v, we  find the final velocity:

v=\frac{mu}{m+M}=\frac{(0.001)(800)}{0.001+1}=0.799 m/s

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3 years ago
A spring with a spring constant value of 125 N/m is compressed 12.2 cm by pushing on it with a 215 g block. When the block is re
allsm [11]

Answer:

v = 2.94 m/s

Explanation:

When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.

Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.

Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means

(1/2)kx^2 = (1/2)mv^2

kx^2 = mv^2

v^2 = (kx^2)/m

v = sqrt((kx^2)/m)

v = x * sqrt(k/m)

v = 0.122 * sqrt(125/0.215)        <--- units converted to m and kg

v = 2.94 m/s

3 0
2 years ago
Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs down
MA_775_DIABLO [31]

Answer:

a)   D_ total = 18.54 m,   b)        v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

           sin 45 = y₁ / d

           cos 45 = x₁ / d

           y₁ = d sin 45

           x₁ = d sin 45

           y₁ = 8 sin 45 = 5,657 m

           x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

            y₂ = 12 m

            x₂ = 0

total displacement

          y_total = y₁ + y₂

          y_total = 5,657 + 12

          y_total = 17,657 m

          x_total = x₁ + x₂

          x_total = 5,657 + 0

          x_total = 5,657 m

          D_total =   17.657 i^+ 5.657 j^  m

          D_total = Ra (17.657 2 + 5.657 2)

          D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

           v = Δx /Δt

the distance traveled using the pythagorean theorem is

         r = √ (d1² + d2²)

          r = √ (8² + 12²)

          r = 14.42 m

The time used for this shredding is

         t = t1 + t2

         t = 1 + 1.2

         t = 2.2 s

let's calculate the average speed

         v = 14.42 / 2.2

         v = 6.55 m / s

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Answer:3764.282 KPa

Explanation:

Given gusher shoots oil at h=25 m

i.e. the velocity of jet is

v=\sqrt{2gh}[/tex]

v=22.147 m/s

Now the pressure loss in pipe is given by hagen poiseuille equation

\Delta P=\frac{32L\mu v}{D^2}

\Delta P=\frac{32\times 50\times 22.147\times 1}{10^{-2}}

\Delta P=3543.557 KPa

For  25 m head in terms of Pressure

\Delta P_2=\rho \times g\times h=220.725 KPa

Total Pressure=\Delta P+\Delta P_2=3543.557+220.725=3764.282 KPa

4 0
3 years ago
A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol
il63 [147K]

Answer:

v = 12.1 m/s

Explanation:

  • When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
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  • Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
  • This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
  • At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

       F_{c} = N + m*g (1)

  • Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

       F_{c} = m*\frac{v^{2}}{r} (2)

  • Since the normal force takes any value as needed to make (1) equal to (2),  if the speed diminishes, it will be needed less force to keep the equality valid.
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  • In this condition, from (1) and (2), we can find the minimum possible value of  the speed that still keeps the motorcycle touching the surface, as follows:
  • v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)
6 0
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