91Ó°ÊÓ

i) Mass of the rider is, m=85 kg.

ii) Initial velocity is,${\mathrm{v}}_{\mathrm{i}}=37\frac{\mathrm{m}}{\mathrm{s}}$.

iii) Acceleration is, ${\mathrm{a}}_1=2.0\frac{\mathrm{m}}{{\mathrm{s}}^2}$.

As we are given the acceleration and mass of the rider, using Newton’s 2^nd law, we can find the force acting on them. As we know the initial velocity, final velocity, and acceleration, we can find the distance traveled. As we get the force and displacement, we can find the work done by the force. Usingthesame procedure, we can find the force, distance, and work done whentherider has an acceleration of $\mathbf{a}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\frac{\mathbf{m}}{{\mathbf{s}}^{\mathbf{2}}}$.

The relationship between force exerted on a particle mass m of the particle and acceleration a of the particle can be written as,

$\mathrm{F}=\mathrm{ma}$ (1)

Substitute the values in the given expression, and we get the magnitude of the force as,

$$\begin{gathered} \left|\stackrel{→}{\mathrm{F}}\right|=\left|85×\left(-2.0\right)\right| \\ \mathrm{F}=\left|-1.7×{10}^2\mathrm{N}\right| \\ \mathrm{F}=1.7×{10}^2\mathrm{N} \end{gathered}$$

Thus, the magnitude of Force when $\mathrm{a}=2.0\mathrm{m}/{\mathrm{s}}^2$is $1.7×{10}^2\mathrm{N}$.

Newton's law of motion, we can write the below expression and calculate the traveled distance x as,

${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_{\mathrm{i}}^2+2\mathrm{ax}$ (2)

As the rider is coming to a stop, so we can say that the final velocity is,
${\mathrm{v}}_{\mathrm{f}}=0\frac{\mathrm{m}}{\mathrm{s}}$

Substitute the values in the given expression, and we get,

$\begin{array}{rcl}0& =& {37}^2-\left(2×2×\mathrm{d}\right)\\ {37}^2& =& 4\mathrm{d}\\ \mathrm{d}& =& \frac{1369}{4}\\ \mathrm{d}& =& 3.4×{10}^2\mathrm{m}\end{array}$

Thus, the distance traveled before coming to a stop when $\mathrm{a}=2.0\frac{\mathrm{m}}{{\mathrm{s}}^2}$is $3.4×{10}^2\mathrm{m}$.

If the force F is exerted on a body and it traveled distance d, then the expression for the work done can be expressed as,

$\mathrm{W}=\mathrm{Fd}$ (3)

Substitute the values in the given expression, and we get,

$$\begin{gathered} \mathrm{W}=\left(-1.7×{10}^2\right)×\left(3.4×{10}^2\right) \\ \mathrm{W}=-5.8×{10}^4\mathrm{J} \end{gathered}$$

Thus, work done by the force on the rider when $\mathrm{a}=2.0\mathrm{m}/{\mathrm{s}}^2$is $-5.8×{10}^4\mathrm{J}$.

Given that,

$\mathrm{a}=4.0\frac{\mathrm{m}}{{\mathrm{s}}^2}$

From equation 1, we can calculatethe magnitude of the force as,

$$\begin{gathered} \left|\stackrel{→}{\mathrm{F}}\right|=\left|85×\left(-4.0\right)\right| \\ \mathrm{F}=\left|\left(-3.4×{10}^2\mathrm{N}\right)\right| \\ \mathrm{F}=3.4×{10}^2\mathrm{N} \end{gathered}$$

Thus, the magnitude of force when data-custom-editor="chemistry" $\mathrm{a}=4.0\mathrm{m}/{\mathrm{s}}^2$is $3.4×{10}^2\mathrm{N}$.

Given that,

$\mathrm{a}=4.0\frac{\mathrm{m}}{{\mathrm{s}}^2}$

From equation 2, we can calculate the distance in this case as,

$\begin{array}{rcl}0& =& {37}^2-\left(2×4×\mathrm{d}\right)\\ {37}^2& =& 8\mathrm{d}\\ \mathrm{d}& =& \frac{1369}{8}\\ \mathrm{d}& =& 1.7×{10}^2\mathrm{m}\end{array}$

Thus, the distance traveled before coming to a stop when $\mathrm{a}=4.0\mathrm{m}/{\mathrm{s}}^2$is $1.7×{10}^2\mathrm{m}$

From equation 3, we can calculate the work done as,

$$\begin{gathered} \mathrm{W}=\left(-3.4×{10}^2\right)×\left(1.7×{10}^2\right) \\ \mathrm{W}=-5.8×{10}^4\mathrm{J} \end{gathered}$$

Thus, work done by the force on the rider when data-custom-editor="chemistry" $\mathrm{a}=4.0\mathrm{m}/{\mathrm{s}}^2$is $-5.8×{10}^4\mathrm{J}$.