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S_A_V [24]
3 years ago
15

Which of the following statements best describes the characteristic of the restoring force in the spring-mass system described i

n the introduction?The restoring force is constant.The restoring force is directly proportional to the displacement of the block.The restoring force is proportional to the mass of the block.The restoring force is maximum when the block is in the equilibrium position
Physics
2 answers:
Grace [21]3 years ago
5 0

Answer : The restoring force is directly proportional to the displacement of the block.

Explanation :

Restoring force is defined as the force that is exerted by the spring due to its mass.

Mathematically, the restoring force can be written as :

F\propto-x

F = - k x

where,

k is the spring constant.

x is the displacement caused due to the mass.

Negative sign shows that the force is acting in opposite direction.

So, it is clear that the restoring force is directly proportional to the displacement of the block.

Hence, the correct option is (b) " The restoring force is directly proportional to the displacement of the block ".

Alex17521 [72]3 years ago
5 0

The correct statement for the answer to the question is <u><em>The style of recovery is directly proportional to the displacement of the block.</em></u>

<h2>Further explanation</h2>

The restoring force is owned by every elastic object that is affected by the force so that the elastic object changes shape. The force that arises in an elastic object to pull back on an object attached to it is called the restoring force. As a result of the restoring force, the object will make a simple harmonic motion. Thus, in objects that perform simple harmonic motions, the restoring force always leads to the object's equilibrium point.

Spring is an example of an elastic object. Because of this elastic nature, a spring that is compressed or stretched will return to its equilibrium state. For example when a spring is given a compressive force, and then the compressive force is removed from the spring, the spring will automatically return to its original shape, from this it can be concluded that the force arising on the spring to restore its position to a state of equilibrium is called the restoring force on the spring.

The magnitude of the restoring force on a spring according to Robert Hooke is formulated as follows.

Fp = -k. Δy

A minus sign indicates that the recovery force is always in the opposite direction to the intersection.

The restoring force on springs is widely used in the fields of engineering and everyday life. For example, in-vehicle and spring shock breakers. Inside the shock breaker, there is a spring that serves to dampen vibrations when the vehicle wheels pass through uneven roads.

Learn more

The restoring force brainly.com/question/11511167, brainly.com/question/6945619

Details

Class: High School

Subject: Physics

Keyword: The restoring force on a spring.

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A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa
Nataly_w [17]

Answer:

V_p = 267.258 m/s

V_n = 38.375 m/s      

Explanation:

using the law of the conservation of the linear momentum:

P_i = P_f

where P_i is the inicial momemtum and P_f is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

P_i = m(270 m/s)

P_f = mV_P + M_nV_n

where m is the mass of the proton and V_p is the velocity of the proton after the collision, M_n is the mass of the nucleus and V_n is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = mV_p + 14mV_n

then, m is cancelated and we have:

270 = V_p + 14V_n

This is a elastic collision, so the kinetic energy K is conservated. Then:

K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2

and

Kf = \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

then,

\frac{1}{2}m(270)^2 =  \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

here we can cancel the m and get:

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

now, we have two equations and two incognites:

270 = V_p + 14V_n  (eq. 1)

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

in the second equation, we have:

36450 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2  (eq. 2)

from this last equation we solve for V_n as:

V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

and replace in the other equation as:

270 = V_p + 14\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

so,

V_p = -267.258 m/s

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

V_n = \frac{270+267.258}{14}

V_n = 38.375 m/s  

3 0
3 years ago
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