# Download YESS 2010 Work and Energy II: Small Group Problems Problem 1

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```YESS 2010
Work and Energy II: Small Group Problems
Problem 1: Escape Velocity
a) Calculate the speed required for a rocket to escape Earth’s gravitational field. Assume the rocket’s
initial position is on the surface of Earth. After the escape, assume that the rocket is infinitely far from
Earth and that its final velocity is approximately zero. (We want to calculate the escape speed necessary
to just barely escape Earth’s gravity.) Does escape speed depend on the mass of the rocket?
√
b) A rocket is accelerated to speed v = 2 gR near Earth’s surface and then it coasts upward. Show that
this rocket will escape from Earth. Determine the speed of the rocket very far from Earth.
SOLUTION:
a) Set up an energy balance. Be sure to use the general form of gravitational potential energy.
∆K + ∆U = 0
1
1
−GM m −GM m
mv22 − mv12 +
−
=0
2
2
r2
r1
From the problem statement: v2 = 0, r1 = R, and r2 = ∞. So we have:
−GM m
1
=0
0 − mv12 + 0 −
2
R
Solve for v1 :
r
v1 = vesc =
2GM
= 11.18 km/s
R
The escape speed does not depend on the mass of the rocket.
p
b) From part (a), we know that a rocket must be accelerated to vesc p
= 2GM/R near Earth’s surface
√
in order to escape. Since g =p
GM/R2 , we have that v2 = 2 gR = 2 GM/R. This is larger than the
required escape speed, vesc = 2GM/R, and therefore, the rocket will escape Earth.
Since the rocket’s initial velocity exceeds the escape velocity, it will have some residual velocity at infinity.
∆K + ∆U = 0
1
−GM m −GM m
1
2
2
mv − mv +
−
=0
2 2 2 1
r2
r1
p
From the problem statement: v1 = 2 GM/R, r1 = R, and r2 = ∞. So we have:
1
2GM m
−GM m
2
mv −
+ 0−
=0
2 2
R
R
Solve for v2 to obtain:
r
v2 =
2GM
R
1
Problem 2: Friction
Consider a block on a plane with coefficient of friction, µ1 , inclined at an angle, θ. Determine its final
speed, v, when it reaches the bottom of the slope, a distance d from its starting point. Solve this problem
two ways: 1) using energy principles, and 2) using force and motion. Compare your results.
SOLUTION USING ENERGY:
∆K + ∆U = WN C
1
1
2
2
mv − mv + (mgh2 − mgh1 ) = WN C
2 2 2 1
Let state 1 correspond to the initial position of the mass at the top of the incline. Let state 2 correspond
to the final position of the mass at the bottom of the incline.
The work due to friction is: Wf = F~f · d~ = −Ff d, where d is the length of the path that the block
traveled. Also recall that Ff = µFN .
Our energy balance becomes:
1
2
mv − 0 + (0 − mgh) = −µ1 FN d
2
1
mv 2 − mgd sin θ + µ1 mgd cos θ = 0
2
Solve the above equation for v:
p
v = 2gd(sin θ − µ1 cos θ)
SOLUTION USING FORCE/MOTION:
Let the positive x direction be along the incline. Set up a force balance:
ΣF~ = m~a
Consider the x component of the force:
Fgx − Ff = max
mg sin θ − µ1 FN = max
mg sin θ − µ1 mg cos θ = max
Solve for ax :
ax = g(sin θ − µ1 cos θ)
To solve for the final speed, use the following kinematic equation:
2
We know that v1 = 0 so we have:
Substitute in for a and solve for v:
p
v = 2gd(sin θ − µ1 cos θ)
The results are the same for both methods as expected.
Problem 3: Air Resistance
In early test flights for the space shuttle using a “glider” (mass of 1000 kg including the pilot), it was
noted that after a horizontal launch at 500 km/h at a height of 3500 m, the glider eventually landed at a
speed of 200 km/h. a) What would the glider’s landing speed have been in the absence of air resistance?
b) What was the average force of air resistance exerted on the glider if it came in at a constant glide angle
of 10◦ to the Earth? (The angle is measured from the horizontal.) Use energy principles to calculate
SOLUTION:
a) Set up an energy balance:
∆K + ∆U = WN C
1
1
−GM m −GM m
mv22 − mv12 +
−
= WN C
2
2
r2
r1
Take r = 0 to be at the center of the Earth. Using this convention, we have that r1 = R + h =
(6378000 + 3500) m = 6381500 m, and r2 =6378000 m. Recall that M is the mass of the Earth.
If there is no air resistance, then WN C = 0. Substitute the values given in the problem. Be sure to
convert speeds from km/h to m/s:
3500 km/h · 1000 m/km
= 138.89 m/s
3600 s/h
v2 = 200 km/h = 55.56 m/s
1 2 1
−(6.67 × 10−11 )(5.9742 × 1024 ) −(6.67 × 10−11 )(5.9742 × 1024 )
2
v − (138.89 ) +
−
=0
2 2 2
6378000
6381500
v1 = 500 km/h =
Solve for the final velocity, v2 :
v2 = 296 m/s = 1066 km/h
b) First determine the work done due to air resistance:
1
1
−GM m −GM m
mv22 − mv12 +
−
= WN C
2
2
r2
r1
1
1000
1000
(1000) 55.562 − 138.892 + (−6.67 × 10−11 )(5.9742 × 1024 )
−
= Wair
2
6378000 6381500
Wair = −4.24 × 107 Nm
The work due to air resistance is given by: Wair = −Fair d.
3
So we have that:
Wair = −Fair
h
sin θ
Solve for force:
sin θ
h
(4.24 × 107 ) sin 10◦
=
3500
Fair = −Wair
Fair
Fair = 2100 N
4
```