Chapter 16.2, Problem 43E

To determine

(a)

To find:

The force acting on the object at time $t$.

Solution:

$F\left(t\right)=2ma\mathbf{i}+6mbt\mathbf{j}$

Explanation:

1) Concept:

The force acting on object of mass $m$ at time $t$ is $F\left(t\right)=mass\times acceleration=m.a\left(t\right)$

2) Given:

$r\left(t\right)=a{t}^2\mathbf{i}+b{t}^3\mathbf{j}, 0\le t\le 1$.

3) Calculations:

The position of an object at time $t$ is

$r\left(t\right)=a{t}^2\mathbf{i}+b{t}^3\mathbf{j}$

Velocity is the rate of change of position with time.

Therefore,

$v\left(t\right)=\frac{dr}{dt}=\frac{d}{dt}\left(a{t}^2\mathbf{i}+b{t}^3\mathbf{j}\right)=2at\mathbf{i}+3b{t}^2\mathbf{j}$

Acceleration is the rate of change of velocity with time.

$a\left(t\right)=\frac{dv}{dt}=\frac{d}{dt}\left(2at\mathbf{i}+3b{t}^2\mathbf{j}\right)=2a\mathbf{i}+6bt\mathbf{j}$

By using the concept,

The force acting on object of mass $m$ at time $t$ is $F\left(t\right)=mass\times acceleration=m.a\left(t\right)$

$F\left(t\right)=m\left(2a\mathbf{i}+6bt\mathbf{j}\right)$

$=2ma\mathbf{i}+6mbt\mathbf{j}$

Conclusion:

The force acting on the object at time $t$ is $F\left(t\right)=2ma\mathbf{i}+6mbt\mathbf{j}$.

(b)

To find:

The work done by the force during the time interval $0\le t\le 1$.

Solution:

$W=2m{a}^2+\frac{9}{2}m{b}^2$

Explanation:

1) Concept:

The work done by the force field is $W={\int }_c^{\square }F·dr={\int }_a^{b}F\left(r\left(t\right)\right)r^{\text{'}}\left(t\right)dt$

2) Given:

$r\left(t\right)=a{t}^2\mathbf{i}+b{t}^3\mathbf{j}, 0\le t\le 1$.

3) Calculation:

From part (a),

$F\left(r\left(t\right)\right)=2ma\mathbf{i}+6mbt\mathbf{j}$

From given,

$r\left(t\right)=a{t}^2\mathbf{i}+b{t}^3\mathbf{j}$

Therefore,

$r^{\text{'}}\left(t\right)=2at\mathbf{i}+3b{t}^2\mathbf{j}$.

By using concept, the work done by the force during the time interval $0\le t\le 1$ is

$W={\int }_c^{\square }F·dr={\int }_0^{1}F\left(r\left(t\right)\right)r^{\text{'}}\left(t\right)dt$

$={\int }_0^1\left(2ma\mathbf{i}+6mbt\mathbf{j}\right)·\left(2at\mathbf{i}+3b{t}^2\mathbf{j}\right)dt$

By using dot product,

$={\int }_0^1\left(4m{a}^{2}t+18m{b}^2{t}^3\right)dt$

By using power rule of integration,

$={\left[\frac{4m{a}^2{t}^2}{2}+\frac{18m{b}^2{t}^4}{4}\right]}_0^1$

Simplify,

$={\left[2m{a}^2{t}^2+\frac{9}{2}m{b}^2{t}^4\right]}_0^1$

Applying limits,

$=\left(2m{a}^2{\left(1\right)}^2+\frac{9}{2}m{b}^2{\left(1\right)}^4\right)-\left(0\right)$

$=2m{a}^2+\frac{9}{2}m{b}^2$

Conclusion:

The work done by the force during the time interval $0\le t\le 1$ is

$W=2m{a}^2+\frac{9}{2}m{b}^2$