Answer to Question #261300 in Physics for jay

Question

A 3.0-kg object has the following two forces acting on it: 

F1 = (16i + 12j) N 

F2 = (−10i + 22j) N 

If the object is initially at rest, determine its velocity v at t = 3.0 s. [INTEGRATE] 

Answer 1

The total force is equal to the sum of the forces (bold font denotes vectors):

$$\mathbf{F} = \mathbf{F_1} + \mathbf{F_2} = (16-10)\mathbf{i}+(12+22)\mathbf{j} = (6\mathbf{i}+34\mathbf{j})N$$

According to the second Newton's law, the acceleration of the object is:

$$\mathbf{a} = \dfrac{\mathbf{F}}{m} =\dfrac{6}{3}\mathbf{i} + \dfrac{34}{3}\mathbf{j} = \left( 2\mathbf{i}+\dfrac{34}{3}\mathbf{j}\right)m/s^2$$

where $m = 3kg$ is the mass of the object.

Integrating, obtain velocity:

$$\mathbf{v} = \int_0^t\mathbf{a}dt = \int_0^t\left( 2\mathbf{i}+\dfrac{34}{3}\mathbf{j}\right)dt =2t \mathbf{i}+\dfrac{34t}{3}\mathbf{j}+C$$

where $t$ is the time and $C$ is the constant of integration (velocity at $t=0$ ). Since the object starts from rest, $C=0$. Substituting $t = 3s$, obtain:

$$\mathbf{v}=2\cdot 3 \mathbf{i}+\dfrac{34\cdot 3}{3}\mathbf{j} = (6\mathbf{i}+34\mathbf{j})m/s$$

Answer. $\mathbf{v}=(6\mathbf{i}+34\mathbf{j})m/s$