Question

A $10.0- kg$ block is released from rest at point (A) in figure. The track is frictionless except for the portion between points (B) and (C), which has a length of $6.00\, m$. The block travels down the track, hits a spring of force constant $3000 \,N / m$ and compresses the spring $0.20\, m$ from its equilibrium position before coming to rest momentarily. If the coefficient of kinetic friction between the. block and the rough surface between points (B) and (C) is $10\, x$. Find the value of $x$.

Answer 1

The easiest way to solve this problem about a chain-reaction process is by considering the energy changes experienced by the block between the point of release (initial) and the point of full compression of the spring (final). Recall that the change in potential energy (gravitational and elastic) plus the change in kinetic energy must equal the work done on the block by non-conservative forces. We choose the gravitational potential energy to be zero along the flat portion of the track.

There is zero spring potential energy in situation (A) and (D) ero gravitational potential energy in situation Putting the energy equation into symbols:
$K_D-K_A-U_{g A}+U_{s D}=-f_k d_{B C}$
Expanding into specific variables:
$0-0-m g y_A+\frac{1}{2} k x_s^2=-f_k d_{B C}$
The friction force is $f_k=\mu_k m g d$,
so $m g y_A-\frac{1}{2} k x^2=\mu_k m g d$
Solving for the unknown variable $\mu_k$ gives
$ \mu_k=\frac{y_A}{d}-\frac{k x^2}{2 m g d}$
$ =\frac{3.00 m }{6.00 m }-\frac{(3000 N / m )(0.20 m )^2}{2(10.0 kg )\left(10 m / s ^2\right)(6.00 m )}=0.40$
Hence, $x=4$.