A body of mass m = 500 kg on an incline of angle α = 30^°…

A body of mass $m = 500\ \mathrm{kg}$ on an incline of angle $\alpha = 30^\circ$ (friction coefficient $\mu = 0.1$) is moving down the incline at $v_0 = 2\ \mathrm{m/s}$. What force $F$ along the incline (pointing up the slope) will stop it in $t_0 = 5\ \mathrm{s}$?

Explanation of Method

Braking a body sliding down an incline.

Step 1: Required deceleration

\[ a = \frac{v_0}{t_0} = \frac{2}{5} = 0.4\ \mathrm{m/s^2} \ \text{(up the slope)}. \]

Step 2: Forces along the incline

Down the slope: $mg\sin\alpha = 2452.5\ \mathrm{N}$. Up the slope: friction $\mu mg\cos\alpha = 424.8\ \mathrm{N}$ and $F$: $F + \mu mg\cos\alpha - mg\sin\alpha = ma$.

\[ F = m\,(a + g\sin\alpha - \mu g\cos\alpha), \]

\[ F = 200 + 2452.5 - 424.8. \]

\[F \approx 2228\ \mathrm{N} \approx 2.23\ \mathrm{kN}\]

$30^\circ$ $F$ $\mu mg\cos\alpha$ $v_0 = 2\ \mathrm{m/s}$ $\text{needed deceleration } \tfrac{2}{5} = 0.4\ \mathrm{m/s^2}$

The block moves down the slope. Friction and the applied force $F$ both point up the slope; together they must exceed the slope component $mg\sin\alpha$ by $m \cdot 0.4\ \mathrm{m/s^2}$.

Result

\[\boxed{\,F \approx 2228\ \mathrm{N} \approx 2.23\ \mathrm{kN}\,}\]

Remark

Check. The original result (2227.6 N) is correct. Once the body has stopped, the force must be reduced (to between 2028 and 2877 N), otherwise it would start pushing the body up the slope. (The mention of a pulley in the statement plays no role.)