Physics, 29.07.2019 17:30 lethycialee79711

Ablock of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7 kg which is initially stationary. the blocks stick together and encounter a rough surface. the blocks eventually come to a stop after traveling a distance d = 1.85 m . what is the coefficient of kinetic friction on the rough surface? μk =

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Answer from: jazwhi1507

First, let's find the speed v_i

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:
p_i = m_1 v_1

After the collision, the two blocks stick together and so now they have mass m_1 +m_2

and they are moving with speed v_i

For conservation of momentum
p_i=p_f

So we can write
m_1 v_1 = (m_1 +m_2)v_i

From which we find
$v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s$

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force $\mu (m_1+m_2)g$ . The work done by the frictional force to stop the two blocks is
$\mu (m_1+m_2)g d$
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
$\frac{1}{2} (m_1+m_2)v_i^2$
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:
$\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d$
From which we can find the value of the coefficient of kinetic friction:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$

Answer from: datboi995

The coefficient of kinetic friction $(\mu)$ is 0.49.

Explanation:

The mass of the block $m_1=3.5\; \text{kg}$

Velocity $V_1=6.3\; \text{m/s}$

The mass of the block $m_2=1.7 \; \text{kg}$

Now, use the conservation of momentum principle:

$p_i=m_1v_1\\p_f=(m_1+m_2)v_i$

we, can write:

m_1v_1=(m_1+m_2)v_2

on putting the values in above equation

we get:

$v_2=\frac{(3.5\times 6.3)}{3.5+1.7}$

$v_2=4.2 \;\text{m/s}$

Now the work done by the frictional force is:

$\mu(m_1+m_2)gd$

Where , d is distance covered by the two blocks.

The initial kinetic energy

$E_i=\frac{1}{2} (m_1+m_2)v_i^2$

Now, the loss in kinetic energy must be equal to work done by friction force:

$\frac{1}{2} (m_1+m_2)v_i^2=\mu(m_1+m_2)gd$

on putting the values in above equation:

$\mu=v_1^2/2gd\\\mu=4.2/(2\times 9.81 \times 1.85)\\\mu=0.49$

Hence , the coefficient of kinetic friction $(\mu)$ is 0.49.

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Answer from: Quest

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Answer from: Quest

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Ablock of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides w...