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Alex
2 years ago
9

A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the

oscillating system is 6.74 J.
A. Determine the following maximum speed of the object?

B. The force constant?

C. The amplitude of the Motion?
Physics
1 answer:
Gnoma [55]2 years ago
4 0

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating  system = 6.74 J

We know that energy is given by

(a) Ke=\frac{1}{2}mv_{max}^2

6.74=\frac{1}{2}\times 0.267\times v_{max}^2

v_{max}=7.1m/sec

(b) Now \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec

We know that \omega =\sqrt{\frac{k}{m}}

35.68=\sqrt{\frac{k}{0.267}}

k=339.9N/m

(c) We know that energy is given by

E=\frac{1}{2}KA^2

6.74=\frac{1}{2}\times 339.9\times A^2

A=19.91cm

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A 20 kg crate initially at rest on a horizontal floor requires a 80 N horizontal force to set it in motion. Find the coefficient
e-lub [12.9K]

Answer:

<em>The coefficient of static friction between the crate and the floor is 0.41</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

Fr=\mu N          [1]

Where \mu is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W = m.g

The crate of m=20 Kg has a weight of:

W = 20*9.8

W = 196 N

The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for \mu:

\displaystyle \mu=\frac{Fr}{N}

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

\displaystyle \mu=\frac{80}{196}

\mu=0.41

The coefficient of static friction between the crate and the floor is 0.41

7 0
2 years ago
A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
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x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0
3 years ago
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