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Parte (b): Calcular la fuerza normal $N$

Step 1: Ecuación de equilibrio vertical (sin aceleración vertical).

$N + P \sin \theta = mg$

Datos: $P = 50~\mathrm{N}$, $\theta = 30^\circ$, $mg = 100~\mathrm{N}$.

$\sin 30^\circ = \frac{1}{2}$

$P \sin \theta = 50 \times \frac{1}{2} = 25~N$

$N = 100 - 25 = 75~N$

$N = 75~\mathrm{N}$

Parte (c): Calcular la aceleración $a$

Step 1: Fuerza de fricción cinética.

$f_k = \mu N = 0.3 \times 75 = 22.5~N$

Step 2: Componente horizontal de $P$.

$\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$

$P \cos \theta = 50 \times 0.866 = 43.3~N$

Step 3: Ecuación de movimiento horizontal ($m = 10~\mathrm{kg}$).

$P \cos \theta - f_k = ma$

$43.3 - 22.5 = 10a$

$20.8 = 10a$

$a = 2.08~m/s^2$

$a = 2.08~\mathrm{m/s^2}$

Parte 2: Si $P$ se duplica ($P=100~\mathrm{N}$), $\mu_s=0.5$, $\mu_k=0.3$, calcular $a$

Step 1: Nueva fuerza normal.

$N + 100 \sin 30^\circ = 100$

$100 \times \frac{1}{2} = 50~N$

$N = 100 - 50 = 50~N$

Step 2: Verificar si se mueve (fricción estática máxima).

$f_{s,\max} = \mu_s N = 0.5 \times 50 = 25~N$

Componente horizontal: $100 \cos 30^\circ = 100 \times 0.866 = 86.6~N > 25~\mathrm{N}$

Sí se mueve (usa fricción cinética).

Step 3: Fuerza de fricción cinética.

$f_k = 0.3 \times 50 = 15~N$

Step 4: Aceleración.

$86.6 - 15 = 10a$

$71.6 = 10a$

$a = 7.16~m/s^2$

$a = 7.16~\mathrm{m/s^2}$