Physics, 25.03.2020 00:36 javontiye226

A block of mass 0.490 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. The force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point circled A, the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The speed of the block at the bottom of the track is vA = 13.4 m/s, and the block experiences an average frictional force of 7.00 N while sliding up the track. What is x? and what is the speed of the block if it reaches the top?

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Answer from: krazyapril4601

The value of x is $x = 0.1955\ m$

The value of the velocity at the top is $v_{top} = 2.25 \ m/s$

Explanation:

The diagram of this process is on the first uploaded image

From the question we are told that

The mass of the block is m_b = 0.490kg

The distance of compression is

The force constant is k = 450 N/m

The radius of the circular track is R =1.00m

The speed at the bottom of the track is $v_A = 13.4 \ m/s$

The frictional force experienced is $F_f = 7.00 \ N$

Now looking at this process we see that the potential energy of the spring is been transformed into the kinetic energy of the block . So,

$PE \ of \ spring = KE \ of \ block$

mathematically i.e

$\frac{1}{2}k x^2 = \frac{1}{2} m_bv^2_A$

0.5 * 450 * x^2 = 0.5 * 0.490 * 13.4^2

Making x the subject of the formula

$x = \sqrt{\frac{0.5 * 0.490 * 13.4^2}{ 0.5 * 450} }$

$= 0.1955\ m$

The average workdone by friction is $W_f = F_f * \pi$

Here $\pi is \ the\ net\ displacement$

W_f = 7 * 3.142

= 21.99 J

The kinetic energy at the bottom is

$KE = \frac{1}{2} * 0.490 * 13.4^2$

$= 43.99 \ N$

The potential energy gained at the top of the circle is

PE = m_bgh

Here h is the height which is equal to d(diameter) = 2r = 2 × 1 = 2 m and

g is acceleration due to gravity = 9.8m/s^2

Now substituting values

PE = 0.490 * 9.8 *2

=9.604 J

Since energy is can not be created nor destroyed but transformed according to first law of thermodynamics

$KE_{at \bottom} = PE _ {\ at \ top } + W_f + KE_{\ at \ top}$

$43.99 = 9.604 +21.99 + [\frac{1}{2} m_b * v_{top} ^2]$

$12.369= [2.45* v_{top} ^2]$

$v_{top} =\sqrt{\frac{12.369}{2.45} }$

$= 2.25 \ m/s$

A block of mass 0.490 kg is pushed against a horizontal spring of negligible mass until the spring i

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