Answer to Question #224670 in Classical Mechanics for Trh

Question

A 4.0kg object has a velocity of 3.0i m/s at one instant. Eight seconds later, its velocity is (8.0i+10.0j) m/s . Assuming the object was subject to a constant net force, find (a) the components of the force and (b) its magnitude.

Answer 1

The given information is

• The initial velocity is $V_{i}=3.0\;i\;\text{m}/\text{s}$
• The final velocity is $V_{f}=(8.0\;i+10.0\;j)\;\text{m}/\text{s}$
• The time it takes to change velocity is $t=8\;\text{s}$
• The mass of the object is $m=4.0\;\text{Kg}$

Part A

The acceleration is given by.

$a=\dfrac{V_{f}-V_{i}}{t}$

where,

$V_{i}$ Is the initial velocity.

$V_{f}$ Is the final velocity.

$t$ Is the time.

Evaluating numerically.

$a=\dfrac{V_{f}-V_{i}}{t}\\ .\\ a=\dfrac{(8.0\;i\;+10.0\;j)\;\text{m}/\text{s}-3.0\;i\;\text{m}/\text{s}}{8.0\;\text{s}}\\ .\\ a=\dfrac{8.0\;i\;\text{m}/\text{s}+10.0\;j\;\text{m}/\text{s}-3.0\;i\;\text{m}/\text{s}}{8.0\;\text{s}}\\ .\\ a=\dfrac{5.0\;i\;\text{m}/\text{s}+10.0\;j\;\text{m}/\text{s}}{8.0\;\text{s}}\\ .\\ a=\dfrac{(5.0\;i+10.0\;j)\;\text{m}/\text{s}}{8.0\;\text{s}}\\ .\\ a=(0.625\;i+1.25\;j)\;\text{m}/\text{s}^{2}\\$

The force on the object is given by.

$F=m\;a\\$

where.

$m$ Is the mass.

$a$ is the acceleration.

Evaluating numerically.

$F=m\;a\\ F=4.0\;\text{Kg}\times (0.625\;i+1.25\;j)\;\text{m}/\text{s}^{2}\\ F=(2.5\;i+5\;j)\;\text{N}$

Answer (a)

The force on the object is 

The horizontal component. $\displaystyle \color{red}{\boxed{F_{x}=2.5\;\text{N}}}$

The vertical component. $\displaystyle \color{red}{\boxed{F_{y}=5\;\text{N}}}$

Part B

The magnitude of the force is given by.

$F=\sqrt{ F_{x}^{2}+F_{y}^{2}}$

Evaluating numerically. 

$F=\sqrt{ F_{x}^{2}+F_{y}^{2}}\\ F=\sqrt{ (2.5\;\text{N})^{2}+(5\;\text{N})^{2}}\\ F=\sqrt{ 31.25\;\text{N}^{2} }\\ F=5.6\;\text{N}$

Answer B

The magnitude of the force on the object is $\displaystyle \color{red}{\boxed{F=5.6\;\text{N}}}$