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2 years ago

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The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is

acted upon by a horizontal force of P=15 lb. Determine the maximum deformation of the outer spring B at the instant the block comes to rest. Spring B has a stiffness of KB=20 lb/ft and the "nested" spring C has a stiffness of kc=40 lb/ft.

1 answer:

Taya2010 [7]2 years ago

4 0

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

W = ΔEm

Where the work is

W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

Em₀ = K = ½ m v²

Final. When the system is at rest

$Em_{f}$ = $K_{e1}b +K_{e2}$ = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

∑ F = F - fr

Axis y

N- W = 0

N = W

The friction force has the equation

fr = μ N

fr = μ W

The job

W = (F – μ W) x

We substitute in the equation

(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values and solve

½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

x² 30 + 14.4 x - 1,125 = 0

x² + 0.48 x - 0.0375 = 0

We solve the second degree equation

x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

x = [-0.48 ± 0.617] / 2

x₁ = 0.0685 m

x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

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a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

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A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

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To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$

Where

A = Cross-sectional Area

Y = Young's modulus

$\alpha$ = Coefficient of linear expansion for steel

$\Delta T$ = Temperature Raise

Our values are given as,

$A = 4.45cm^2$

$T = 37K$

$\alpha = 1.17*10^{-5}K^{-1}$

$Y = 200*10^9Gpa$

Replacing we have,

$F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)$

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Answer:

The airplane should release the parcel $6.7*10^2$ m before reaching the island

Explanation:

The height of the plane is $y_0=100m$ , and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

$y=y_0 - \frac{gt^2}{2}$ [1]

And the distance X is

x = V.t [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

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$x = 150 m/s \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}$

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Answer:

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Explanation:

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The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.

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