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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00 x 102 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00 x 102 N force for an hour.

The required solution is 3.89 x 106 C.

See the step by step solution

## Step 1: The given data

The emf of the battery is: $\mathcal{E}=12.0\mathrm{V}.$

The mass of the car is: $m=750kg$

The car’s initial velocity is: ${\mathrm{v}}_{\mathrm{i}}=\mathrm{0}$

The car’s final velocity is: ${V}_{f}=25.0m/s$

The height of the hill is: $h=2.00×{10}^{2}m$

The force applied on the car is: $F=5.00×{10}^{2}N$

The time for which force is exerted: $∆t=3600s$

## Step 2: Principle and concepts

The kinetic energy of an object is

${\mathbf{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{mv}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(1\right)}}$

where m is the mass of the object and v is its speed relative to the chosen coordinate system.

The potential energy of any object

${\mathbf{U}}{\mathbf{=}}{\mathbf{mgy}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(2\right)}}$

where m is the mass of the object, g=9.80 N/kg, and y is the height

The work formula is given by,

${\mathbf{W}}{\mathbf{=}}{\mathbf{Fdcos\theta }}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(}}{\mathbf{3}}{\mathbf{\right)}}$

where F is the magnitude of the force, d is the magnitude of the displacement, and ${\theta }$ is the angle between the direction of F and the direction of d . The sign ${\mathrm{Cos\theta }}$ determines the sign of the work.

## Step 3: Principle and concepts

The average speed of a particle :

${{\mathbf{v}}}_{{\mathbf{avg}}}{\mathbf{=}}\frac{\mathbf{d}}{\mathbf{\Delta t}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(4\right)}}$

Where d is the total distance and role="math" localid="1654898190581" ${∆}{\mathrm{t}}$ is the total travel time.

The electric potential energy:

${{\mathbf{U}}}_{{\mathbf{e}}}{\mathbf{=}}{\mathbf{q}}{\mathbf{V}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{\left(5\right)}}$

Where q is the charge of particle, V is the potential.

## Step 4: Calculation of kinetic energy

The kinetic energy acquired by the car from equation (1):

${\mathbf{\Delta K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{mv}}}^{{\mathbf{2}}}$

Substitute the values:

$\begin{array}{rcl}{\mathbf{∆}}{\mathbf{K}}& {\mathbf{=}}& \frac{\mathbf{1}}{\mathbf{2}}\left(750\mathrm{kg}\right){\left(25.0m/s\right)}^{{\mathbf{2}}}\\ & {\mathbf{=}}& {\mathbf{234375}}{\mathbf{J}}\end{array}$

The required kinetic energy is 234375J.

## Step 5: Calculation of gravitational potential energy

The gravitational potential energy from equation (2):

${{\mathbf{\Delta U}}}_{{\mathbf{g}}}{\mathbf{=}}{\mathbf{mgh}}$

Evaluating the values

$\begin{array}{rcl}{{\mathbf{\Delta U}}}_{{\mathbf{g}}}& {\mathbf{=}}& {\mathbf{\left(}}{\mathbf{750}}{\mathbf{}}{\mathbf{kg}}{\mathbf{\right)}}\left(9.80m/{s}^{2}\right)\left(2.00×{10}^{2}m\right)\\ & {\mathbf{=}}& {\mathbf{1}}{\mathbf{.}}{\mathbf{47}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{6}}}{\mathbf{}}{\mathbf{J}}\\ & & \end{array}$

The required potential energy is 1.47 x 106 J.

## Step 6: Calculation of Work done

The work done from equation (3):

${\mathbf{W}}{\mathbf{=}}{\mathbf{Fd}}$

where d the distance moved by the car is $\mathrm{v}∆\mathrm{t}$ as found from equation (4):

${\mathbf{W}}{\mathbf{=}}{\mathbf{Fv\Delta t}}$

Substitute the values:

$\begin{array}{rcl}{\mathbf{W}}& {\mathbf{=}}& \left(5.00×{10}^{2}N\right){\mathbf{\left(}}{\mathbf{25}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}{\mathbf{\right)}}{\mathbf{\left(}}{\mathbf{3600}}{\mathbf{}}{\mathbf{s}}{\mathbf{\right)}}\\ & {\mathbf{=}}& {\mathbf{4}}{\mathbf{.}}{\mathbf{50}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{7}}}{\mathbf{}}{\mathbf{J}}\\ & & \end{array}$

The required work done is 4.50 x 107 J.

## Step 7: Calculation of kinetic energy

The car is accelerated by the battery’s energy, making the car climb up the hill (giving it gravitational potential energy as well as make the car travel at a constant speed. So,

$\begin{array}{rcl}{\mathbf{\Delta E}}& {\mathbf{=}}& {\mathbf{\Delta K}}{\mathbf{+}}{{\mathbf{\Delta U}}}_{{\mathbf{g}}}{\mathbf{+}}{\mathbf{W}}\\ & {\mathbf{=}}& {\mathbf{234375}}{\mathbf{}}{\mathbf{J}}{\mathbf{+}}{\mathbf{1}}{\mathbf{.}}{\mathbf{47}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{6}}}{\mathbf{}}{\mathbf{J}}{\mathbf{+}}{\mathbf{4}}{\mathbf{.}}{\mathbf{50}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{7}}}{\mathbf{}}{\mathbf{J}}\\ & {\mathbf{=}}& {\mathbf{46704375}}{\mathbf{}}{\mathbf{J}}\\ & & \end{array}$

## Step 8: Calculation of the charge

According to equation (5), the energy given by the battery is-

${\mathbf{\Delta E}}{\mathbf{=}}{\mathbf{qE}}$

Solving for q:

${\mathbf{q}}{\mathbf{=}}\frac{\mathbf{\Delta E}}{\mathbf{\epsilon }}$

Substitute all values

$\begin{array}{rcl}{\mathbf{q}}& {\mathbf{=}}& \frac{\mathbf{46704375}\mathbf{}\mathbf{J}}{\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}}\\ & {\mathbf{=}}& {\mathbf{3}}{\mathbf{.}}{\mathbf{89}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{6}}}{\mathbf{}}{\mathbf{C}}\\ & & \end{array}$

Therefore, the charge, 3.89 x 106 C battery must supply.