An 80 kg skateboarder moving at 3 m/s pushes off with her back foot to move faster. If her velocity increases to 5 m/s, what is
2 answers:
1) The kinetic energy of an object is given by:
where m is the object's mass and v its speed.
By using this equation, we find the initial kinetic energy of the skateboarder:
and the final kinetic energy as well:
So, her change in kinetic energy is
2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:
Therefore, the work done by the skateboarder is
where m is the object's mass and v its speed.
By using this equation, we find the initial kinetic energy of the skateboarder:
and the final kinetic energy as well:
So, her change in kinetic energy is
2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:
Therefore, the work done by the skateboarder is
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Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s
By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s
Position #2:
Radius, r₂ = 2 ft
By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²
Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s
At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf
The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.
The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb
Answer: 44.25 ft-lb
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s
By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s
Position #2:
Radius, r₂ = 2 ft
By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²
Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
= (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s
At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf
The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.
The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb
Answer: 44.25 ft-lb
a) 4F0
b) Speed of planet B is the same as speed of planet A
Speed of planet C is twice the speed of planet A
Explanation:
a)
The magnitude of the gravitational force between two objects is given by the formula
where
G is the gravitational constant
m1, m2 are the masses of the 2 objects
r is the separation between the objects
For the system planet A - Star A, we have:
So the force is
For the system planet B - Star B, we have:
So the force is
So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.
For the system planet C - Star C, we have:
So the force is
So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.
b)
The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:
where
m is the mass of the planet
M is the mass of the star
v is the tangential speed
We can re-arrange the equation solving for v, and we find an expression for the speed:
For System A,
So the tangential speed is
For system B,
So the tangential speed is
So, the speed of planet B is the same as planet A.
For system C,
So the tangential speed is
So, the speed of planet C is twice the speed of planet A.
Initially, the energies are:
At final point, the energies are:
Using conservation law of energy,
The equation is further simplified as: