Question

A force F = 3j +4j newton is applied on a body of mass 2 kg calculate the magnitude and direction of the acceleration in the body.

Answer 1

Answer: The magnitude of the acceleration is $|a| =2.5 m/s^{2}$ and the direction is $\theta =tan^{-1}\dfrac{4}{3}$.

Explanation:

Given that,

Mass m = 2 kg

Force F = 3i+4j

Using newton's second law

F = ma

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

$a = \dfrac{3i+4j}{2}$

$a = \dfrac{3i}{2}+2j$

$a = 1.5i+2j m/s^{2}$

The magnitude of the acceleration is

$|a|= \sqrt{1.5^{2}+2^{2}}$

$|a| =2.5 m/s^{2}$

Now, the direction of the acceleration is

$\theta =tan^{-1}\dfrac{y}{x}$

$\theta =tan^{-1}\dfrac{4}{3}$

Hence, the magnitude of the acceleration is $|a| =2.5 m/s^{2}$ and the direction is $\theta =tan^{-1}\dfrac{4}{3}$.

Answer 2

"Given:

Force = 3i + 4j

Mass = 2 kg

Solution:

According to Newton's second law of motion,

Force = Mass \times Acceleration

$(3 i+4 j)=2 \times a \rightarrow(1)$

Magnitude of the Force $=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}$

Magnitude of Force = 5 units        

Substitute in (1),

5=2a

$a=\frac{5}{2}=2.5 \frac{m}{s^{2}}$

Direction of the acceleration using the below given formula,

$\theta=\tan ^{-1}\left(\frac{b}{a}\right)$

Where, a=3; b=4

$\theta=\tan ^{-1}\left(\frac{4}{3}\right)$

$\theta=53.1^{\circ}$

The force is 53.1° with positive x-axis. The force is in first quadrant."