How much work is required to lift an object with a mass of 5.0 kilograms to a height of 3.5 meters?
2 answers:
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Answer:
f = 3.09 Hz
Explanation:
This is a simple harmonic motion exercise where the angular velocity is
w² =
to find the constant (k) of the spring, we use Hooke's law with the initial data
F = - kx
where the force is the weight of the body that is hanging
F = W = m g
we substitute
m g = - k x
k =
we calculate
k =
k = 3.769 10² m
we substitute in the first equation
w² =
w = 19.415 rad / s
angular velocity and frequency are related
w = 2πf
f =
f = 19.415 / 2pi
f = 3.09 Hz
Answer:
See the answers below
Explanation:
To solve this problem we must use the definition of power and work in physics.
a)
The function of the conveyor belt is to carry the boxes from an initial point that is at low altitude to an end point that is at high altitude. In this way the conveyor belt prints a speed to the box to be able to raise it to the required vertical distance.
Since we have a velocity at the beginning and then we place the box at a high position, where then the box remains at rest, we can say that it converts kinetic energy to potential energy.
b)
Power is defined as the relationship of work over time. Therefore we have:
where:
P = power = 3.8 [kW] = 3800 [W]
W = work [J]
t = time = 14 [s]
c)
Since the given time is equal to the given time at Point b, we can use the same work calculated.
We know that work is defined as the product of force by the distance traveled.
So, the force is equal to:
Now we know that force is defined as the product of mass by gravitation acceleration.
where:
F = force or weight = 10037.73 [N]
g = gravity acceleration = 9.81 [m/s²]
m = mass [kg]
d)
This part can be solved by means of the energy conservation theorem, where the potential energy is transformed into kinetic energy or vice versa.
where:
h = elevation = 4.7 [m]
v = velocity [m/s]