Is much work does a an elevator do to lift a 1500 kg elevator a height of 100 m?
1 answer:
Work done is equal to the force times the distance over which that force acts (provided they are parallel vectors) In symbols we say:
Since gravity acts downward and the elevator goes upward, the dot product is reduced to a simple multiplication problem. First we need the downward force of the elevator, which is the weight according to F=ma. F=1500kg*9.81m/s^2=14715N. Then W=14715N*100m=1471500J, or 1.4715MJ
Since gravity acts downward and the elevator goes upward, the dot product is reduced to a simple multiplication problem. First we need the downward force of the elevator, which is the weight according to F=ma. F=1500kg*9.81m/s^2=14715N. Then W=14715N*100m=1471500J, or 1.4715MJ
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Answer:
6.0 m/s
Explanation:
According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.
Therefore, we can write:
or
where:
m is the mass of the athlete
u is the initial speed of the athlete (at the bottom)
0 is the initial potential energy of the athlete (at the bottom)
v = 0.80 m/s is the final speed of the athlete (at the top)
is the acceleration due to gravity
h = 1.80 m is the final height of the athlete (at the top)
Solving the equation for u, we find the initial speed at which the athlete must jump:
Answer:
Explanation:
Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.
Due to the definition of cross product, the magnitude of the torque is given by:
Where is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when
is equal to one, solving for r: