Physics, asked by vinitlatehat75, 10 months ago

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​

Answers

Answered by αmαn4чσu
67

 \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{$F_f = 2500N$}}

Answered by ShivamKashyap08
47

\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}

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