Question

if it requires 25 kilowatt to drive an automobile at 36 km per hour along a level road what is the frictional force opposing the motion​

Answer 1

$\large{\textbf{\underline{Frictional\: force \:exerted\: by\: road = 250000 N}}}$

$\huge{\textbf{\underline{As per Question:-}}}$

Power required = 25 kilowatt

Velocity of automobile = 36 km/hr

10 m/s

$\huge{\textbf{\underline{To find :-}}}$

Frictional force exerted by road.

$\huge{\textbf{\underline{Solution:-}}}$

$\large{\textbf{\underline{Power:-}}}$

The work done by an object per unit time is known as power.

$P = \dfrac{W}{T}$

But ,

Work :- When a Force displace a body in direction of force then It is known as work done by force.

$\large{\textbf{ W = FS }}$

• Put the value of W.

$P = \dfrac{F \times S}{T}$

$\huge\textbf{ P = FV }$

Frictional force exerted by road f.

$25000 = \dfrac{F_f}{10}$

$\therefore \textbf{\underline{ F_f = 2500N }}$

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Power required = 25 Kilo watt.
• Speed of the automobile = 36 Km/h.
• Frictional force = f.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Let's see the definition of Power,

Power is defined as rate of doing work.

Its units are:-

• S.I units = Watt.
• C.G.S units = erg/second.
• Commercial unit :- KWH.[Kilowatt hour]

Its Formula is given by:-

$\large{\boxed{\tt P = \dfrac{W}{t}}}$

Here ,

• P denotes Power,
• W denotes Work done.
• t denotes Time period.

For, Instantaneous Power,

$\large{\bold{\tt P = \dfrac{dW}{dt}}}$

[Taking over a short interval of time]

Now,

W = F.S

Substituting it,

$\large{\tt P = \dfrac{F.S}{dt}}$

As Force is constant it cannot be differentiated.

$\large{\tt P = F . \left[ \dfrac{dS}{t} \right]}$

As we know, ${\tt \dfrac{ds}{dt} = velocity}$

Substituting it,

$\large{\boxed{\tt P = F.V}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, From the question,

Velocity is given in Km/hr, it needs to be converted to m/s.

Therefore,

$\large{\tt V = 36 \: km/hr = 36 \times \dfrac{5}{18}}$

$\large{\tt V = 36 \: km/hr = \cancel{36} \times \dfrac{5}{\cancel{18}}}$

Therefore, Speed will be,

$\large{\tt V = 10 \: m/s \: \: ----(1)}$

Now,

Applying the formula of instantaneous Power.

$\large{\boxed{\tt P = F.V}}$

Substituting the values,

$\large{\tt 25,000 = f \times 10 \: \: \: \: \: \: \: [25KW = 25,000 \: W]}$

$\large{\tt f = \dfrac{25,000}{10}}$

Now,

$\large{\tt f = \dfrac{25,00\cancel{0}}{\cancel{10}}}$

$\huge{\boxed{\boxed{\tt f = 2,500 \: N}}}$

So, the Frictional force opposing the motion is 2,500 N.

$\rule{300}{1.5}$