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Tju [1.3M]
4 years ago
8

1. A 500-g block is placed on a level, frictionless surface and attached to an ideal spring. At t = 0 the block moves through th

e equilibrium position with speed vo in the –x direction, as shown below. At t =  sec, the block reaches its maximum displacement of 40 cm to the left of equilibrium. a. Determine the value of each of the following quantities. Show all work. • period: • spring constant: b. Using x(t) = A cos(t + o) as the solution to the differential equation of motion: i. Determine the form of the function v(t) that represents the velocity of the block. ii. Evaluate all constant parameters (A, , and o) so as to completely describe both the position and velocity of the block as functions of time. c. Consider the following statement about the situation described above. "It takes the first  seconds for the block to travel 40 cm, so the initial speed vo can be found by dividing 40 cm by  seconds." Do you agree or disagree with this statement? If so, explain why you agree. If not, explain why you disagree and calculate the initial speed vo of the block.
Physics
1 answer:
Thepotemich [5.8K]4 years ago
6 0

Answer:

Pls refer to attached handwritten document for calculation and detailed answer

Explanation:

You might be interested in
Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

  •  Solid Sphere I = 2/5* m *r²

                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

  • Cylinder I = 1/2 * m* r²

                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

5 0
3 years ago
A skater has rotational inertia 4.2 kg.m2 with his fists held to his chest and 5.7 kg.m2 with his arms outstretched. The skater
Fynjy0 [20]

Answer:

6.13428 rev/s

Explanation:

I_f = Final moment of inertia = 4.2 kgm²

I = Moment of inertia with fists close to chest = 5.7 kgm²

\omega_i = Initial angular speed = 3 rev/s

\omega_f = Final angular speed

r = Radius = 76 cm

m = Mass = 2.5 kg

Moment of inertia of the skater is given by

I_i=I+2mr^2

In this system the angular momentum is conserved

L_f=L_i\\\Rightarrow I_f\omega_f=I_i\omega_i\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{(5.7+2\times 2.5\times 0.76^2)3}{4.2}\\\Rightarrow \omega_f=6.13428\ rev/s

The rotational speed will be 6.13428 rev/s

5 0
3 years ago
A 75.0-kg person climbs stairs, gaining 2.50 m in height. Find the work done to accomplish this task.
ololo11 [35]

Answer:

W=1837.5J

Explanation:

A force exerts work when there is a displacement of its point of application in the direction of that force. Therefore, the work done by a system is defined as the inner product between the applied force and the displacement:

W=\vec{F}\cdot \vec{d}\\W=Fcos\theta d

In this case, we have:

F=mg\\h=dcos\theta

So, replacing this:

W=mgh\\W=75kg*9.8\frac{m}{s^2}*2.5m\\W=1837.5J

4 0
4 years ago
Un movil avanza a 20 m/s y recorre una distancia de 800 km. Determinar el tiempo en horas que utiliza
Nataly_w [17]

Answer:

t = 11.1 hours

Explanation:

The question says that, "A mobile advances at 20 m / s and travels a distance of 800 km. Determine the time in hours you use".

Given that,

Speed of a mobile, v = 20 m/s = 72 km/h

Distance, d = 800 km

We know that,

Speed = distance/time

So,

t=\dfrac{d}{v}\\\\t=\dfrac{800}{72}\\\\t=11.1\ h

So, it will take 11.1 hours.

7 0
3 years ago
A car with a mass of 1,500 kg is traveling at a speed of 30 m/s. What force must be applied to stop the car in 3 seconds?​
QveST [7]

Explanation:

F = ma

F = m Δv / Δt

F = (1500 kg) (0 m/s − 30 m/s) / 3 s

F = -15,000 N

3 0
3 years ago
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