The lift does 3000 J of work in 5 seconds. What is the power of the lift?

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretch

r-ruslan [8.4K]

Answer:

a) A = 0.98 m

b) Ф = 90°

c) x = -0.98sin(12.25t)

Explanation:

We know the value of the spring constant which is 300 N/m, the innitial apmplitude, that we will call it xo is 0, at t = 0, and the speed is 12 m/s

The expression for the amplitude under these conditions is:

A = √xo² + vx²/w² (1)

To calculate the angular speed w, we use the following expression:

w = √k/m (2)

Calculating w:

w = √300/2 = 12.25 rad/s

Now, we replace this value into equation 1, along with the other known values and solve for A:

A = √0 + (12)²/(12.25)²

A = 0.98 m

b) In this part, is actually easy, the displacement of x in function of the time is given by:

x = A cos(wt - Ф) (4)

But at t = 0 we have then:

x = xo = A cosФ (5)

Solving for the angle Ф we have:

xo/A = cosФ

Ф = arccos(x0/A) (6)

Replacing the data in (6):

Ф = arccos(0/0.98)

Ф = 90°

c) Equation (4) is the expression for the simple harmonic motion

x = A cos(wt - Ф)

And if we replace here the value of w and the previous angle, we can write an equation for x in function of t:

x = 0.98 cos (12.25t - 90)

x = 0.98 cos(12.25t - 90)

And we have an trigonometric expression for cos that is:

cos(α - π/2) = -sinα

in this case, α will be the value of w = 12.25 rad/s. and 90° is the same as rewritting as π/2, therefore:

cos(α - π/2) = cos(12.25t - 90)

x = -0.98sin(12.25t)

3 0

3 years ago

A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius

disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

KE = 0.23 J

given parameters

Disk radius R = 15 cm = 0.15 m
Cylinder radius r = 1.5 cm = 0.0015 m
Disk mass M = 20 kg
Time t = 1.2 s
Force F = 12 N

to find

Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk I_{disk} = ½ M R²

cylinder I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

I_{total} = ½ (M R² + m r²) = ½ M R² ( $1 + \frac{m}{M} \ (\frac{r}{R})^2$ )