Ask question

Ask question

All categories

4 years ago

8

A spool of thin wire (with inner radius r = 0.45 m, outer radius R = 0.60 m, and moment of inertia Icm = 0.8208 kg·m2) pivots on

a shaft. The wire is pulled down by a mass M = 1.300 kg. After falling a distance D = 0.500 m, starting from rest, the mass has a speed of v = 63.1 cm/s. Calculate the energy lost to friction during that time.

1 answer:

makkiz [27]4 years ago

3 0

Answer:

$E_L=4.667J$

Explanation:

The energy lost can be model by the energy equation in potential and kinetic energy in each step of the motion

Given: $r=0.45m$ , $R=0.60m$ , $I_m=0.8208 kg*m^2$ , $v=63.1cm/s$ , $m=1.3kg$

$E_p=m*g*r$

$E_p=1.3kg*9.8m/s^2*0.45m$

$E_p=5.733 J$

Kinetic energy:

$K_{Es}=\frac{1}{2}*I*w^2$ , $K_{Ep}=\frac{1}{2}*m*v^2$

$w=v^2/r$

Energy lost :

$E_p-K_{Es}-K_{Ep}=0$

$5.733J-\frac{1}{2}*0.8208kg*(\frac{(0.631m/s)^2}{0.45m})^2-\frac{1}{2}*1.3kg*(0.631m/s)^2=0$

$E_L=4.667J$

You might be interested in

A stone is thrown vertically into the air at an initial velocity of 96 ft/s. On Mars, the height s (in feet) of the stone above

vladimir1956 [14]

Answer:

240 ft

Explanation:

t = Time taken

u = Initial velocity = 96 ft/s

v = Final velocity

s = Displacement

a = Acceleration = 12 m/s² on Mars 32 ft/s² on Earth negative due to upward direction

Mars

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -12\times t^2\\\Rightarrow s=96t-6t^2\ ft$

Earth

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -32\times t^2\\\Rightarrow s=96t-16t^2\ ft$

Differentiating the first equation with respect to time we get

$\frac{ds}{dt}=96-12t$

Equating with zero

$0=96-12t\\\Rightarrow t=\frac{96}{12}=8\ s$

Differentiating the second equation with respect to time we get

$\frac{ds}{dt}=96-32t$

Equating with zero

$0=96-32t\\\Rightarrow t=\frac{96}{32}=3\ s$

Applying the time taken to the above equations, we get

$s=96t-6t^2\ ft\\\Rightarrow s=96\times 8-6\times 8^2\\\Rightarrow s=384$

$s=96t-16t^2\\\Rightarrow s=96\times 3-16\times 3^2\\\Rightarrow s=144$

Difference in height = 384-144 = 240 ft

The stone will travel 240 ft higher on Mars

6 0

3 years ago

When water is in a container the surface of the water is curved what the curve called?

Alex777 [14]

It is called the meniscus.

3 0

4 years ago

Where is the magnitude of the magnetic field around a permanent magnet greatest?

Nata [24]

Near the poles as the lines are the closest

5 0

3 years ago

It takes 40 J of energy to push a large box 4m across a floor. Assuming the push is the same direction as the movement, what is

Jobisdone [24]

Answer:

10 N

Explanation:

Work done is the dot product of the force magnitude with distance and cosine of the angle betweenthem.

Equation to use: W=|F|*|d|*cosθ

Your unknown is the force F; You have one equation, oneunknown.

40j=(F)*(4m)*cos(0)

F=40j/4m

F=10N

3 0

3 years ago

Read 2 more answers

(eText prob. 4.25 with some values changed) Air enters a diffuser of a jet engine operating at steady state at 2.65 psia, 389◦R,

krek1111 [17]

Answer:

$V_2 = 45.44m/s$

Explanation:

We have to many data in different system, so we need transform everything to SI, that is

$P_1 = 2.65 Psi = 18.271 kPa\\T_1= 389\°R = 216 K\\V_1 = 869ft/s = 264m/s\\T_2 = 450\°R = 250K$

When we have all this values in SI apply a Energy Balance Equation,

$\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0$

Solving for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

From the table of gas properties we calculate for $T_1 = 216K$ and $T_2 = 250K$

$h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})$

$h_1 = 215.97kJ/kg$

For T_2;

$h_2 = 250.05kJ/kg$

Substituting in equation for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

$V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}\\V_2 = 45.44m/s$

4 0

3 years ago