Math / Calculus

QuestionThe 20.0 W electric motor of a toy car is ranked as $91.0 \%$ efficient. If the motor is to accelerate the 740 g car from rest along a horizontal surface, then determine the maximum speed that the car can attain in 3.80 seconds.

Studdy Solution

STEP 1

1. The motor's power output is $20.0 \, \text{W}$ .
2. The motor efficiency is $91.0\%$ .
3. The car's mass is $740 \, \text{g}$ or $0.740 \, \text{kg}$ .
4. The car starts from rest.
5. The car moves along a horizontal surface.
6. The time duration for acceleration is $3.80 \, \text{s}$ .

STEP 2

1. Calculate the useful power output of the motor.
2. Determine the work done by the motor.
3. Use the work-energy principle to find the final kinetic energy.
4. Calculate the maximum speed of the car.

STEP 3

Calculate the useful power output of the motor using its efficiency:
$\text{Useful Power} = \text{Efficiency} \times \text{Total Power}$ $= 0.91 \times 20.0 \, \text{W}$

STEP 4

Calculate the work done by the motor over $3.80 \, \text{s}$ :
$\text{Work} = \text{Useful Power} \times \text{Time}$ $= (0.91 \times 20.0 \, \text{W}) \times 3.80 \, \text{s}$

STEP 5

Use the work-energy principle, which states that the work done is equal to the change in kinetic energy:
$\text{Work} = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$
Since the car starts from rest, $u = 0$ , so:
$\text{Work} = \frac{1}{2} m v^2$

STEP 6

Solve for the maximum speed $v$ :
$\frac{1}{2} m v^2 = \text{Work}$ $v^2 = \frac{2 \times \text{Work}}{m}$ $v = \sqrt{\frac{2 \times \text{Work}}{m}}$
Substitute the values:
$v = \sqrt{\frac{2 \times (0.91 \times 20.0 \, \text{W} \times 3.80 \, \text{s})}{0.740 \, \text{kg}}}$
Calculate $v$ .
The maximum speed of the car is:
$\boxed{9.27 \, \text{m/s}}$

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QuestionThe 20.0 W electric motor of a toy car is ranked as $91.0 \%$ efficient. If the motor is to accelerate the 740 g car from rest along a horizontal surface, then determine the maximum speed that the car can attain in 3.80 seconds.

Studdy Solution

STEP 1

1. The motor's power output is $20.0 \, \text{W}$ .
2. The motor efficiency is $91.0\%$ .
3. The car's mass is $740 \, \text{g}$ or $0.740 \, \text{kg}$ .
4. The car starts from rest.
5. The car moves along a horizontal surface.
6. The time duration for acceleration is $3.80 \, \text{s}$ .

STEP 2

1. Calculate the useful power output of the motor.
2. Determine the work done by the motor.
3. Use the work-energy principle to find the final kinetic energy.
4. Calculate the maximum speed of the car.

STEP 3

Calculate the useful power output of the motor using its efficiency:
$\text{Useful Power} = \text{Efficiency} \times \text{Total Power}$ $= 0.91 \times 20.0 \, \text{W}$

STEP 4

Calculate the work done by the motor over $3.80 \, \text{s}$ :
$\text{Work} = \text{Useful Power} \times \text{Time}$ $= (0.91 \times 20.0 \, \text{W}) \times 3.80 \, \text{s}$

STEP 5

Use the work-energy principle, which states that the work done is equal to the change in kinetic energy:
$\text{Work} = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$
Since the car starts from rest, $u = 0$ , so:
$\text{Work} = \frac{1}{2} m v^2$

STEP 6

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QuestionThe 20.0 W electric motor of a toy car is ranked as 91.0%91.0 \%91.0% efficient. If the motor is to accelerate the 740 g car from rest along a horizontal surface, then determine the maximum speed that the car can attain in 3.80 seconds.

Studdy Solution

STEP 1

STEP 2

1. Calculate the useful power output of the motor.2. Determine the work done by the motor.3. Use the work-energy principle to find the final kinetic energy.4. Calculate the maximum speed of the car.

STEP 3

Calculate the useful power output of the motor using its efficiency:Useful Power=Efficiency×Total Power \text{Useful Power} = \text{Efficiency} \times \text{Total Power} Useful Power=Efficiency×Total Power =0.91×20.0 W = 0.91 \times 20.0 \, \text{W} =0.91×20.0W

STEP 4

Calculate the work done by the motor over 3.80 s 3.80 \, \text{s} 3.80s:Work=Useful Power×Time \text{Work} = \text{Useful Power} \times \text{Time} Work=Useful Power×Time =(0.91×20.0 W)×3.80 s = (0.91 \times 20.0 \, \text{W}) \times 3.80 \, \text{s} =(0.91×20.0W)×3.80s

STEP 5

STEP 6

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionThe 20.0 W electric motor of a toy car is ranked as $91.0 \%$ efficient. If the motor is to accelerate the 740 g car from rest along a horizontal surface, then determine the maximum speed that the car can attain in 3.80 seconds.

1. Calculate the useful power output of the motor.
2. Determine the work done by the motor.
3. Use the work-energy principle to find the final kinetic energy.
4. Calculate the maximum speed of the car.

Calculate the useful power output of the motor using its efficiency:
$\text{Useful Power} = \text{Efficiency} \times \text{Total Power}$ $= 0.91 \times 20.0 \, \text{W}$

Calculate the work done by the motor over $3.80 \, \text{s}$ :
$\text{Work} = \text{Useful Power} \times \text{Time}$ $= (0.91 \times 20.0 \, \text{W}) \times 3.80 \, \text{s}$