Question

1. two forces act on a 250 kg mass, 150 n e and 350 n s 45° w. calculate the acceleration of the mass.

Explanation:

Step1: Resolve the second force into components

The second force $F_2 = 350\ N$ at $S45^{\circ}W$.
The west - east component of $F_2$ is $F_{2x}=- 350\sin45^{\circ}\ N$ and the north - south component is $F_{2y}=-350\cos45^{\circ}\ N$.
$F_{2x}=-350\times\frac{\sqrt{2}}{2}\approx - 247.49\ N$ and $F_{2y}=-350\times\frac{\sqrt{2}}{2}\approx - 247.49\ N$. The first force $F_1 = 150\ N$ (east).
The net force in the x - direction $F_{netx}=150 - 247.49=- 97.49\ N$.
The net force in the y - direction $F_{nety}=-247.49\ N$.

Step2: Calculate the magnitude of the net force

Using the Pythagorean theorem, $F_{net}=\sqrt{F_{netx}^{2}+F_{nety}^{2}}$.
$F_{net}=\sqrt{(-97.49)^{2}+(-247.49)^{2}}=\sqrt{9504.4 + 61252.3}=\sqrt{70756.7}\approx266.01\ N$.

Step3: Calculate the acceleration

According to Newton's second law $F = ma$, where $m = 250\ kg$.
$a=\frac{F_{net}}{m}=\frac{266.01}{250}=1.064\ m/s^{2}$.

Answer:

$1.06\ m/s^{2}$ (rounded to two decimal places)