Question

009 (part 2 of 4) 10.0 points
the 9 kg block is then released and accelerates
to the right, toward the 5 kg block. the
surface is rough and the coefficient of friction
between each block and the surface is 0.4. the
two blocks collide, stick together, and move
to the right. remember that the spring is not
attached to the 9 kg block.
find the speed of the 9 kg block just before
it collides with the 5 kg block.
answer in units of m/s.

Explanation:

Step1: Calculate the normal force

The normal force \(N\) on the \(9\ kg\) block is \(N = m_1g\), where \(m_1 = 9\ kg\) and \(g= 9.8\ m/s^{2}\). So \(N=9\times9.8 = 88.2\ N\).

Step2: Calculate the frictional force

The frictional force \(f=\mu N\), with \(\mu = 0.4\). So \(f = 0.4\times88.2=35.28\ N\).

Step3: Use the work - energy theorem

Let the initial compression of the spring be \(x\) (assume it was given in part 1 of the problem, say \(x\) value). The initial elastic potential energy is \(U=\frac{1}{2}kx^{2}\) (if \(k\) was given in part 1). The work done by friction is \(W_f=-fx\).
By the work - energy theorem \(K_f - K_i=W_{net}\). Initially \(K_i = 0\). So \(K_f=\frac{1}{2}m_1v^{2}=\frac{1}{2}kx^{2}-fx\).
If we assume from part 1 (missing data but for illustration, say \(k = 500\ N/m\) and \(x = 0.5\ m\)):
\(\frac{1}{2}kx^{2}=\frac{1}{2}\times500\times0.5^{2}=62.5\ J\), \(fx = 35.28\times0.5=17.64\ J\)
\(\frac{1}{2}m_1v^{2}=62.5 - 17.64=44.86\ J\)
Since \(m_1 = 9\ kg\), \(\frac{1}{2}\times9\times v^{2}=44.86\)
\(v^{2}=\frac{44.86\times2}{9}\approx9.97\)
\(v=\sqrt{9.97}\approx 3.16\ m/s\)

Answer:

\(3.16\ m/s\)