A block is initially at position x = 0 and in contact with an uncompressed spring of negligible mass. The block is pushed back a
long a frictionless surface from position x = 0 to x = -D , as shown above, compressing the spring by an amount Δx = D . The block is then released. At x = 0 the block enters a rough part of the track and eventually comes to rest at position x = 3D . The coefficient of kinetic friction between the block and the rough track is µ.
On the axes below, sketch and label graphs of the following two quantities as a function of the position of the block between x = -D and x = 3D . You do not need to calculate values for the vertical axis, but the same vertical scale should be used for both quantities.
i. The kinetic energy K of the block
ii. The potential energy U of the block-spring system
On the axes below, sketch and label graphs of the following two quantities as a function of the position of the block between x = -D and x = 3D . You do not need to calculate values for the vertical axis, but the same vertical scale should be used for both quantities.
i. The kinetic energy K of the block
ii. The potential energy U of the block-spring system
1 answer:
You might be interested in
Answer:
(a) W = 1329.5 J = 1.33 KJ
(b) ΔU = 24.27 KJ
Explanation:
(a)
Work done by the gas can be found by the following formula:
where,
W = Work = ?
P = constant pressure = (0.991 atm)() = 100413 Pa
ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)() = 0.01324 m³
Therefore,
W = (100413 Pa)(0.01324 m³)
<u>W = 1329.5 J = 1.33 KJ</u>
<u></u>
(b)
Using the first law of thermodynamics:
ΔU = ΔQ - W (negative W for the work done by the system)
where,
ΔU = change in internal energy of the gas = ?
ΔQ = heat added to the system = 25.6 KJ
Therefore,
ΔU = 25.6 KJ - 1.33 KJ
<u>ΔU = 24.27 KJ</u>
Answer:
mechanical energy per unit mass is 887.4 J/kg
power generated is 443.7 MW
Explanation:
given data
average velocity = 3 m/s
rate = 500 m³/s
height h = 90 m
to find out
total mechanical energy and power generation potential
solution
we know that mechanical energy is sum of potential energy and kinetic energy
so
E = ×m×v² + m×g×h .............1
and energy per mass unit is
E/m = ×v² + g×h
put here value
E/m = ×3² + 9.81×90
E/m = 887.4 J/kg
so mechanical energy per unit mass is 887.4 J/kg
and
power generated is express as
power generated = energy per unit mass ×rate×density
power generated = 887.4× 500× 1000
power generated = 443700000
so power generated is 443.7 MW
<h2>Answer:</h2>
<em>Hello, </em>
<h3><u>QUESTION)</u></h3>According to the second Newton's Law,
<em>✔ We have : F = m x a ⇔ m = F/a </em>
- m = 3000/15
- m = 200 kg
The mass of the object is therefore 200 kg.