A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 14.9 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 3h0 above the spring, what would be the maximum compression of the spring?

Answers

Answer 1

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= [tex]\frac{1}{2}[/tex]kx² = [tex]\frac{1}{2}[/tex](530)(0.149)² =  5.88 J

b)  Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg[tex]h_o[/tex] = [tex]\frac{1}{2}[/tex]kx² - mgx

And, [tex]h_o[/tex]= ( [tex]\frac{1}{2}[/tex]kx² - mgx )/(mg) = [tex][\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)[/tex]]/(0.645x9.8)    

   [tex]h_o[/tex]=   0.78m            

d) Now, if the initial initial height of block is 3[tex]h_o[/tex]

[tex]h_o[/tex] = 3 x 0.78 = 2.34m

then, [tex]\frac{1}{2}[/tex]kx² - mgx - mg[tex]h_o[/tex] =0

 

[tex]\frac{1}{2}[/tex](530)x²  - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0  

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum  compression of the spring)


Related Questions

A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an efficiency of 30%, and taken in 50 kJ from the hot steam per cycle. If a Carnot engine takes in the same amount of heat per cycle and operates at these temperatures, the work it can turn into is most likely to be:a) 15 kJ. b) 20 kJ. c) 10 kJ. d) 50 kJ.

Answers

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁  Where T₁ is temperature of hot source  and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .

To understand thermal linear expansion in solid materials. Most materials expand when their temperatures increase. Such thermal expansion, which is explained by the increase in the average distance between the constituent molecules, plays an important role in engineering. In fact, as the temperature increases or decreases, the changes in the dimensions of various parts of bridges, machines, etc., may be significant enough to cause trouble if not taken into account. That is why power lines are always sagging and parts of metal bridges fit loosely together, allowing for some movement. It turns out that for relatively small changes in temperature, the linear dimensions change in direct proportion to the temperature.
For instance, if a rod has length L0 at a certain temperature T0 and length L at a higher temperature T, then the change in length of the rod is proportional to the change in temperature and to the initial length of the rod: L - L0 = αL0(T - T0),
or
ΔL = αL0ΔT.
Here, α is a constant called the coefficient of linear expansion; its value depends on the material. A large value of α means that the material expands substantially as the temperature increases; smaller values of α indicate that the material tends to retain its dimensions. For instance, quartz does not expand much; aluminum expands a lot. The value of α for aluminum is about 60 times that of quartz!
Questions:
A) Compared to its length in the spring, by what amount ΔLwinter does the length of the bridge decrease during the Teharian winter when the temperature hovers around -150°C?
B) Compared to its length in the spring, by what amount ΔLsummer does the length of the bridge increase during the Teharian summer when the temperature hovers around 700°C?

Answers

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)

Answers

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

[tex]v(t)=ate^{-6t}[/tex]   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]

hence, the maximum speed is v_max = ((1/6)e^-1)a

A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?

Answers

Answer:

-7.23 rad/s

Explanation:

Given that

Mass of the cockroach, m = 0.157 kg

Radius of the disk, r = 14.9 cm = 0.149 m

Rotational Inertia, I = 5.92*10^-3 kgm²

Speed of the cockroach, v = 2.92 m/s

Angular velocity of the rim, w = 3.89 rad/s

The initial angular momentum of rim is

Iw = 5.92*10^-3 * 3.89

Iw = 2.3*10^-2 kgm²/s

The initial angular momentum of cockroach about the axle of the disk is

L = -mvr

L = -0.157 * 2.92 * 0.149

L = -0.068 kgm²/s

This means that we can get the initial angular momentum of the system by summing both together

2.3*10^-2 + -0.068

L' = -0.045 kgm²/s

After the cockroach stops, the total inertia of the spinning disk is

I(f) = I + mr²

I(f) = 5.92*10^-3 + 0.157 * 0.149²

I(f) = 5.92*10^-3 + 3.49*10^-3

I(f) = 9.41*10^-3 kgm²

Final angular momentum of the disk is

L'' = I(f).w(f)

L''= 9.41*10^-3w(f)

Using the conservation of total angular momentum, we have

-0.068 = 9.41*10^-3w(f) + 0

w(f) = -0.068 / 9.41*10^-3

w(f) = -7.23 rad/s

Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed

b)

The mechanical energy of the cockroach is not converted as it stops

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