Question

Only two horizontal forces act on a 2.0 kg body. One force is 3.2 N, acting due east, and the other is 7.1 N, acting 38° north of west. What is the magnitude of the body's acceleration?

Answer 1

Answer

given,

mass of the body = 2 kg

force acting on east = 3.2 N

Force acting = 7.1 at 38° north of west

$\vec{F_1} = 3.2 \hat{i} N$

$\vec{F_2} = - 7.1 cos 38^0 \hat{i}+8.5 sin 38^0 \hat{j}$

                = $-5.59 \hat{i} + 5.23 \hat{j}$

$F_net = F_1 + F_2$

$F_net = 3.2 \hat{i} + -5.59 \hat{i} + 5.23 \hat{j}$

$F_net = -2.39 \hat{i} + 5.23 \hat{j}$

magnitude of F

$F = \sqrt{(-2.39)^2+5.23^2}$

F = 5.75 N

$tan \theta = \dfrac{5.23}{-2.39}$

$\theta = tan^{-1}(-2.19)$

$\theta = 65.46$

$a = \dfrac{F}{m}$

$a = \dfrac{5.75}{2}$

a = 2.875 m/s²