If a boat and its riders have a mass of 900 kg and the boat drifts in at 1.4 m/s how much work does sam do to stop it?
1 answer:
The initial kinetic energy of the boat and its rider is
After Sam stops it, the final kinetic energy of the boat+rider is
because its final velocity is zero.
For the law of conservation of energy, the work done by Sam is the variation of kinetic energy of the system:
where the negative sign is due to the fact that the force Sam is applying goes against the direction of motion of the boat.
After Sam stops it, the final kinetic energy of the boat+rider is
because its final velocity is zero.
For the law of conservation of energy, the work done by Sam is the variation of kinetic energy of the system:
where the negative sign is due to the fact that the force Sam is applying goes against the direction of motion of the boat.
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Answer:
A) = 1.44 kg m², B) moment of inertia must increase
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is
I = ½ m R²
A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is
I = + m D²
Let's apply these equations to our case
The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms
=
+ 2
= ½ M R²
The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body
M = 7/8 m total
M = 7/8 64
M = 56 kg
The mass of the arms is
m’= 1/8 m total
m’= 1/8 64
m’= 8 kg
As it has two arms the mass of each arm is half
m = ½ m ’
m = 4 kg
The arms are very thin, we will approximate them as a particle
= M D²
Let's write the equation
= ½ M R² + 2 (m D²)
Let's calculate
= ½ 56 0.20² + 2 4 0.20²
= 1.12 + 0.32
= 1.44 kg m²
b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase