Physics, 19.09.2019 16:30 Braxton12

Block a of mass ma is moving horizontally with speed va along a frictionless sur- face. it collides with block b of mass mb that is initially at rest. the two blocks stick together after the collision. at x = 0, block b enters a rough surface with a coefficient of kinetic friction that increases linearly with distance μk(x) = bx for 0 ≤ x ≤ d, where b is a positive constant. at x = d, block b collides with an unstretched spring with spring constant k on a frictionless surface. the downward gravitational acceleration has magnitude g. what is the distance the spring is compressed when the blocks first comes to rest? express you answer in terms of some or all of the following: va, b,d, g,k, ma and mb.

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Answer from: sloane50

The distance the spring is compressed when the blocks first come to rest is: $\mathbf{\sqrt{ \dfrac{m_A^2v_A^2-(m_A+m_B) ^2bgd^2}{k(m_A+m_B)}}}$

From the information given, using the conservation of linear momentum on both blocks A and B, we have:

$\mathbf{m_Av_A + 0 = (m_A+m_B)V}$

$\mathbf{V = \dfrac{m_Av_A}{m_A+m_B}}$

here;

V represents the velocity of the two blocks after collision

If we look at the frictional force on the fused two blocks after sticking torether, we get:

$\mathbf{f= \mu _kN}$

where;

coefficient of kinetic friction is $\mu_k$ = bxN = normal reaction

The normal reaction N can be expressed by the formula:

$\mathbf{= (m_A +m_B) g}$

replacing the formula for the normal reaction into $\mathbf{f= \mu _kN}$

$\mathbf{f =bx ( m_A+m_B ) g}$

$\mathbf{f = ( m_A+m_B )bx g}$

However, the work done by the frictional force is:

$\mathbf{dW = fdx}$

$\mathbf{dW =( m_A+m_B )bx g dx}$

So, if we can estimate the net work done at interval 0 ≤ x ≤ d, we will be a step closer to determining the distance that the spring is compressed at the time the block first comes to rest.

So;

Net work done $\mathbf{W =\int ^d _o (m_A +m_B)bxgdx}$

$\mathbf{ \implies (m_A+m_B) bg \Big [ \dfrac{x^2}{2}\Big]^d_o}$

$\mathbf{ \implies \dfrac{(m_A+m_B)bgd^2}{2}}$

Similarly, the energy of the blocks prior to getting into the frictional force region is:

$\mathbf{E = \dfrac{1}{2}(m_A + m_B) V^2}}$

$\mathbf{\implies \dfrac{1}{2}(m_A+m_B) \Big [\dfrac{m_Av_A}{m_A+m_B} \Big] ^2}$

$\mathbf{\implies\dfrac{m_A^2v_A^2}{2(m_A+m_B)} }$

Finally, the distance(x) at which the spring is compressed when the block first comes to rest can be estimated by determining the difference between the energy of the two blocks and the net work done;

Using the relation:

$\mathbf{E_{spring} = E -W}}$

$\mathbf{\dfrac{1}{2}kx^2= \dfrac{m_A^2v_A^2}{2(m_A+m_B)}- \dfrac{(m_A+m_B) bgd^2}{2}}$

$\mathbf{kx^2= \dfrac{m_A^2v_A^2-(m_A+m_B) ^2bgd^2}{(m_A+m_B)}}$

$\mathbf{x^2= \dfrac{m_A^2v_A^2-(m_A+m_B) ^2bgd^2}{k(m_A+m_B)}}$

$\mathbf{x=\sqrt{ \dfrac{m_A^2v_A^2-(m_A+m_B) ^2bgd^2}{k(m_A+m_B)}}}$

Therefore, we can conclude that the distance the spring is compressed when the blocks first come to rest is $\mathbf{\sqrt{ \dfrac{m_A^2v_A^2-(m_A+m_B) ^2bgd^2}{k(m_A+m_B)}}}$

Learn more about work done here:

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Answer from: paras29

Th spring is compressed by distance $x=\sqrt{\dfrac{1}{k} \left (\dfrac{m_A^2v_v_A^2}{(m_A+m_B)}-(m_A+m_B)bgd^2 \right )}\\$

Explanation:

Given:

Mass of block A= m_A

Mass of block B= m_B

variation of coefficient of friction with x, $\mu_k(x)=bx$

spring constant=k

Distance covered on frictional surface=d

Taking block A and block B as a system , there is no external force acting on the system so the momentum can b conserved in horizontal direction.

Conservation Of Momentum

$m_av_A=(m_A+m_B)v\\v=\dfrac{m_av_A}{m_A+m_B}$

Now the blocks got stick together so both of them will pass through the frictional surface and will compress the spring together.

Work done by friction

$W_f=-\int \mu gb (m_A+m_B)x\ dx\\W_f=\dfrac{(m_A+m_B)bgd^2}{2}$

Work done by spring

$W_s=-\dfrac{kx^2}{2}$

So applying Work Energy Theorem before the blocks moves to the frictional surface and when the blocks comes to rest by co pressing the spring by distance x.

Work done by all the spring +work done by friction=change in kinetic Energy of the system of blocks.

$\dfrac{-(m_A+m_B)bgd^2}{2}-\dfrac{kx^2}{2}=0-\dfrac{(m_A+m_B)V^2}{2}$

$x=\sqrt{\dfrac{1}{k} \left (\dfrac{m_A^2v_A^2}{(m_A+m_B)}-(m_A+m_B)bgd^2 \right )}$

Block a of mass ma is moving horizontally with speed va along a frictionless sur- face. it collides

Answer from: Quest

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

Answer from: Quest

The reflection of the candle in the concave mirror formed an upside-down virtual image that is smaller than the candle.in other words. the answer is the third option, c