An 8.00-kg block of ice, released from rest at the top of a 1.50-m long frictionless ramp, slides downhill, reaching a speed of 2.50 m/s at the bottom. (a) What is the angle between the ramp and the horizontal? (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.0 N parallel to the surface of the ramp?

The constant opposing friction force is $f=10\mathrm{N}$

The mass of block is $m=8\mathrm{kg}$

The length of ramp is $I=1.5\mathrm{m}$

The speed of block at the bottom of ramp is $v=2.5\mathrm{m}/\mathrm{s}$

The speed of block at the top of ramp is $u=0$

The work done by force to move block from top of ramp to bottom is equal to the change in kinetic energy of block between top and bottom of ramp.

When the body is sliding on the inclined plane then the friction force is called kinetic friction and it is denoted by ${\mathit{f}}_{\mathbf{k}}$.

(a)

Draw the diagram of the system,

The angle of the ramp from horizontal is calculated as by using work energy theorem:

$mgl\sin \theta =\frac{1}{2}\left(v^2-u^2\right)$

Here, is the angle of the ramp from horizontal, $u$is the initial speed of block and its value is zero due to rest position of block, $v$is the final speed of the block, $g$is the acceleration due to gravity and m is the mass of the ice block.

Substitute 8kg for $m$, $9.8\mathrm{m}/{\mathrm{s}}^2$ for $g$, $1.5\mathrm{m}$for $l$, $2.50\mathrm{m}/\mathrm{s}$ for $v$and 0m/s for $u$ in the above equation.

$$\begin{gathered} \left(8kg\right)\left(9.8m/s^2\right)\left(1.5m\right)\sin \theta =\frac{1}{2}\left(8kg\right)\left[{\left(2.50m/s\right)}^2-{\left(0\right)}^2\right] \\ \sin \theta =0.213 \\ \theta =12.3° \end{gathered}$$

Therefore, the angle of ramp from horizontal is $12.3°$.

(b)

$\left(mg\sin \theta -f\right)l=\frac{1}{2}\left(V^2-u^2\right)$

Here, $f$is the constant friction force, $V$is the speed of block at the bottom of ramp.

Substitute 8kg for $m$, $9.8\mathrm{m}/{\mathrm{s}}^2$ for g , 1.5 m for $l$ , 2.50m/s for v and 0m/s for u and 10N for $f$in the above equation.

$$\begin{gathered} \left[\left(8\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\sin 12.3°\right)-10\mathrm{N}\right]\left(1.5\mathrm{m}\right)=\frac{1}{2}\left(8\mathrm{kg}\right)\left[{\mathrm{V}}^2-{\left(0\right)}^2\right] \\ V=1.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of block at bottom with friction force is $1.6\mathrm{m}/\mathrm{s}$.