Question

A block of mass 1.5 kg is held against a light spring compressed to 60cm from its equilibrium length of 80cm. the spring constant k = 600 n/m. all surfaces are smooth except a 25-cm long rough patch bc with co-efficient of kinetic friction μk = 0.4. the block, after being released, moves over a smooth surface before entering the rough patch bc and finally moving up the incline to a maximum height h. [a] how fast is the block moving as it enters the rough patch at b ? [b] (8 pts) how fast is the block moving as it exits the rough patch at c ? [c] what is the value of h ?

Answer 1

A) when the spring is released, all the elastic potential energy of the spring is converted into kinetic energy of the mass in motion. So we can write:
$\frac{1}{2} k (\Delta x)^2 = \frac{1}{2} m v^2$
where $k=600~N/m$ is the elastic constant of the spring, $\Delta x = 80~cm-60~cm=20~cm=0.2~m$ is its displacement, $m=1.5~kg$ is the mass of the block and $v$ the velocity of the block after it is released from the spring. So we can find v:
$v= \sqrt{ \frac{k (\Delta x)^2}{m} } = 4~m/s$
This is the velocity the block has before entering the rough path bc.

b) After it enters the rough path, the block starts to lose energy because of the work $W_{fr}$ done by the frictional force with frictional coefficient of $\mu=0.4$. So, calling Kf and Ki the final and initial kinetic energy, we have
$K_f = K_i - W_{fr}$
the length of the rough path bc is 20 cm=0.2 m, so the work done by the frictional force is 
$W=\mu mg d=1.47~J$
while the initial kinetic energy is
$K_i = \frac{1}{2} m v_i^2 = 12~J$
Therefore kinetic energy after the rough path is
$K_f=12~J-1.47~J=10.53~J$
and the velocity is
$v_f= \sqrt{ \frac{2 K_f}{m} } = 3.75~m/s$

c) $K_f$ found at point b) is the kinetic energy of the block before entering the incline. At the maximum height h, all this kinetic energy is converted into potential energy, mgh. So we can write
$K_f = mgh$
and from this we find h:
$h= \frac{K_f}{mg} =0.72~m$