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zlopas [31]
2 years ago
13

An unstretched spring has a length of 10. centimeters. When the spring is stretched by a force of 16 newtons, its length is incr

eased to 18 centimeters. What is the spring constant of this spring
Physics
1 answer:
Fed [463]2 years ago
5 0

Answer:

The spring constant of this spring  is 200 N/m.

Explanation:

Given:

Original unstretched length of the spring (x₀) = 10 cm =0.10 m [1 cm =0.01 m]

Stretched length of the spring (x₁) = 18 cm = 0.18 cm

Force acting on the spring (F) = 16 N

Spring constant of the spring (k) = ?

First let us find the change in length of the spring or the elongation caused in the spring due to the applied force.

So, Change in length = Final length - Initial length

\Delta x = x_1-x_0=0.18-0.10=0.08\ m

Now, restoring force acting on the spring is directly related to its elongation or compression as:

F=k\Delta x

Rewriting in terms of 'k', we get:

k=\dfrac{F}{\Delta x}

Now, plug in the given values and solve for 'k'. This gives,

k=\frac{16\ N}{0.08\ m}\\\\k=200\ N/m

Therefore, the spring constant of this spring  is 200 N/m.

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Answer:

  • v_1  =  \ 5.196 \frac{m}{s}
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Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

\vec{p}_i = \vec{p}_f

where the suffix i  means initial, and the suffix f means final.

The initial momentum will be:

\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}

as the second puck is initially at rest:

\vec{v}_{2_i} = 0

Using the unit vector \vec{i} pointing in the original line of motion:

\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}

\vec{p}_i = 0.70 \ kg  \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0

\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}

So:

\vec{p}_i =  4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f

\vec{p}_f =  4.2 \ \frac{kg \ m}{s} \ \hat{i}

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

So, our velocity vectors will be:

\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )

\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

We got

\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}

4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \   v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )  + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

So, we got the equations:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg \   v_1 \  cos(30 \°) + 0.7 \ kg \ v_2 \  cos(-60 \°)

and

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0  = 0.7 \ kg \  ( v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°) )

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v_1 \  sin(30 \°) = -  \ v_2 \  sin(-60 \°)

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and, for the first one:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg  \ (  v_1 \  cos(30 \°) + v_2 \  cos(60 \°) )

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} =    (\ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} = v_2     (\   \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) +   cos(60 \°)

6 \ \frac{m}{s} = v_2  * 2

so:

v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}

and

v_1  =  \ 3 \frac{m}{s}  \  \frac{sin(60 \°)}{ sin(30 \°) }

v_1  =  \ 5.196 \frac{m}{s}

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