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Question: A motor of \[100\,{\text{H}}{\text{.P}}{\text{.}}\] is moving with a constant velocity of \[72\,{\te...

A motor of 100H.P.100\,{\text{H}}{\text{.P}}{\text{.}} is moving with a constant velocity of 72km/hour72\,{\text{km/hour}}. The forward force exerted by the engine on the car is
A. 3.73×103N{\text{3}}{\text{.73}} \times {\text{1}}{{\text{0}}^3}\,{\text{N}}
B. 3.73×102N{\text{3}}{\text{.73}} \times {\text{1}}{{\text{0}}^2}\,{\text{N}}
C. 3.73×101N{\text{3}}{\text{.73}} \times {\text{1}}{{\text{0}}^1}\,{\text{N}}
D. None of the above

Explanation

Solution

Use the expression relating between the power, force and velocity. Convert the final answer for the forward force exerted by the engine on the car in the power of 10.

Formula used:
The relation between the power, force and velocity is
P=FvP = Fv …… (1)
Here, PP is the power, FF is the force and vv is the velocity.

Complete step by step answer:
The power PP of the motor of the engine is 100H.P.100\,{\text{H}}{\text{.P}}{\text{.}} and the velocity vv of the motor is 72km/hour72\,{\text{km/hour}}.
P=100H.P.P = 100\,{\text{H}}{\text{.P}}{\text{.}}
v=72km/hourv = 72\,{\text{km/hour}}
Convert the unit of the power of the motor from horsepower to watt.
P=(100H.P.)(745.7W1H.P.)P = \left( {100\,{\text{H}}{\text{.P}}{\text{.}}} \right)\left( {\dfrac{{745.7\,{\text{W}}}}{{1\,{\text{H}}{\text{.P}}{\text{.}}}}} \right)
P=74570W\Rightarrow P = 74570\,{\text{W}}
Convert the unit of the velocity of the car from kilometer per hour to meter per second.
v=(72kmhour)(103m1km)(1hour3600s)v = \left( {72\,\dfrac{{{\text{km}}}}{{{\text{hour}}}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{m}}}}{{1\,{\text{km}}}}} \right)\left( {\dfrac{{1\,{\text{hour}}}}{{3600\,{\text{s}}}}} \right)
v=20m/s\Rightarrow v = 20\,{\text{m/s}}
Rearrange equation (1) for the forward force exerted by the engine on the car.
F=PvF = \dfrac{P}{v}
Substitute 74570W74570\,{\text{W}} for PP and 20m/s20\,{\text{m/s}} for vv in the above equation.
F=74570W20m/sF = \dfrac{{74570\,{\text{W}}}}{{20\,{\text{m/s}}}}
F=3728.5Wm1s\Rightarrow F = 3728.5\,{\text{W}} \cdot {{\text{m}}^{ - 1}} \cdot {\text{s}}
F=3728.5N\Rightarrow F = 3728.5\,{\text{N}}
F3.73×103N\Rightarrow F \approx 3.73 \times {10^3}\,{\text{N}}
Hence, the forward force exerted by the engine on the car is 3.73×103N3.73 \times {10^3}\,{\text{N}}.
Hence, the correct option is A.

Note: Use proper conversion factors to convert the units of the power and the velocity. The units of all the physical quantities substituted in the formula for the force must be in the SI system of units.Also remember that from Newton’s second law of motion,the force is directly proportional to the rate of change of momentum of a body.