Question

The figure(Figure 1) shows a 1.75-kg block at rest on a ramp of height h. When the block is released, it slides without friction to the bottom of the ramp, and then continues across a surface that is frictionless except for a rough patch of width 10.0 cm that has a coefficient of kinetic friction 0.650. Part A Find h such that the block's speed after crossing the rough patch is 3.40 m/s .

Answer 1

At the top of the ramp (1) , the block has gravitational potential energy

$\small P_{E1}=mgh$

then, when it reaches the rough surface (2), it only has kinetic energy:

$\small K_{E2}=\frac{1}{2}m{v_2}^2$

and also, after the rough surface (3), the block has only kinetic energy:

$\small K_{E3}=\frac{1}{2}m{v_3}^2$

From 1 to 2, there are only conservative forces actuating on the block, so

$\small P_{E1}=K_{E2}$

$\small 1.75kg\times 9.81m/s^2\times h=\frac{1}{2}1.75kg\times {v_2}^2$

$\small v_2=\sqrt{2gh}$

Then, from 2 to 3, we have

$\small W_{friction}=\Delta K_E$

$\small -\mu mg\times d=\frac{1}{2}m{v_3}^2-\frac{1}{2}m{v_2}^2$

$\small -0.65\times 9.81m/s^2\times 0.10m=\frac{1}{2}{(3.40 m/s)}^2-\frac{1}{2}\left ({\sqrt{2\times9.81m/s^2\times h }} \right )^2$

solving for h,

$\small -0.63765\: m^2/s^2={5.78\: m^2/s^2}-9.81m/s^2\times h$

$\small h =\frac{{5.78\: m^2/s}^2+0.63765\: m^2/s^2}{9.81m/s^2 }$

$\small h =0.654\: m$