A motor must lift a 1000-kg elevator cab. The cab’s maximum occupant capacity is 400 kg, and its constant "cruising" speed is 1.
1 answer:
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Answer:
Explanation:
Here we will call:
1. : The energy when the first spring is compress
2. : The energy after the mass is liberated by the spring
3. : The energy before the second string catch the mass
4. : The energy when the second sping compressed
so, the law of the conservations of energy says that:
1.
2.
3.
where is the work of the friction.
1. equation 1 is equal to:
where K is the constant of the spring, x is the distance compressed, M is the mass and the velocity, so:
Solving for velocity, we get:
= 65.319 m/s
2. Now, equation 2 is equal to:
where M is the mass, the velocity in the situation 2,
is the velocity in the situation 3,
is the coefficient of the friction, N the normal force and d the distance, so:
Volving for , we get:
3. Finally, equation 3 is equal to:
where is the constant of the second spring and
is the compress of the second spring, so:
solving for , we get:
Answer:
h = 5.05 m
Explanation:
given,
mass of pole vaulter, m = 64 Kg
speed of the vaulter,V = 10.2 m/s
horizontal component of velocity in air, v = 1 m/s
height of the jump,h = ?
using energy conservation
initial height of the vaulter is equal to zero.
h = 5.05 m
height of the jump is equal to 5.05 m.