Respuesta :
First, let's find the speed [tex]v_i[/tex] of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
[tex]p_i = m_1 v_1[/tex]
After the collision, the two blocks stick together and so now they have mass [tex]m_1 +m_2[/tex] and they are moving with speed [tex]v_i[/tex]:
[tex]p_f = (m_1 + m_2)v_i[/tex]
For conservation of momentum
[tex]p_i=p_f[/tex]
So we can write
[tex]m_1 v_1 = (m_1 +m_2)v_i[/tex]
From which we find
[tex]v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s [/tex]
The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force [tex]\mu (m_1+m_2)g[/tex]. The work done by the frictional force to stop the two blocks is
[tex]\mu (m_1+m_2)g d[/tex]
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
[tex] \frac{1}{2} (m_1+m_2)v_i^2 [/tex]
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
[tex] \frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d[/tex]
From which we can find the value of the coefficient of kinetic friction:
[tex]\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49 [/tex]
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
[tex]p_i = m_1 v_1[/tex]
After the collision, the two blocks stick together and so now they have mass [tex]m_1 +m_2[/tex] and they are moving with speed [tex]v_i[/tex]:
[tex]p_f = (m_1 + m_2)v_i[/tex]
For conservation of momentum
[tex]p_i=p_f[/tex]
So we can write
[tex]m_1 v_1 = (m_1 +m_2)v_i[/tex]
From which we find
[tex]v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s [/tex]
The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force [tex]\mu (m_1+m_2)g[/tex]. The work done by the frictional force to stop the two blocks is
[tex]\mu (m_1+m_2)g d[/tex]
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
[tex] \frac{1}{2} (m_1+m_2)v_i^2 [/tex]
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
[tex] \frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d[/tex]
From which we can find the value of the coefficient of kinetic friction:
[tex]\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49 [/tex]
The coefficient of kinetic friction [tex](\mu)[/tex] is 0.49.
Explanation:
The mass of the block [tex]m_1=3.5\; \text{kg}[/tex]
Velocity [tex]V_1=6.3\; \text{m/s}[/tex]
The mass of the block [tex]m_2=1.7 \; \text{kg}[/tex]
Now, use the conservation of momentum principle:
[tex]p_i=m_1v_1\\p_f=(m_1+m_2)v_i[/tex]
we, can write:
[tex]m_1v_1=(m_1+m_2)v_2[/tex]
on putting the values in above equation
we get:
[tex]v_2=\frac{(3.5\times 6.3)}{3.5+1.7}[/tex]
[tex]v_2=4.2 \;\text{m/s}[/tex]
Now the work done by the frictional force is:
[tex]\mu(m_1+m_2)gd[/tex]
Where , d is distance covered by the two blocks.
The initial kinetic energy
[tex]E_i=\frac{1}{2} (m_1+m_2)v_i^2[/tex]
Now, the loss in kinetic energy must be equal to work done by friction force:
[tex]\frac{1}{2} (m_1+m_2)v_i^2=\mu(m_1+m_2)gd[/tex]
on putting the values in above equation:
[tex]\mu=v_1^2/2gd\\\mu=4.2/(2\times 9.81 \times 1.85)\\\mu=0.49[/tex]
Hence , the coefficient of kinetic friction [tex](\mu)[/tex] is 0.49.
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