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AP Physics 1 : Power

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Example Question #11 : Power

Which of the following units best represents power?

Possible Answers:

$\frac{N\cdot m}{s}$

$kg\cdot \frac{m}{s^2}$

$kg\cdot \frac{m}{s}$

$\frac{N\cdot kg}{s}$

Correct answer:

$\frac{N\cdot m}{s}$

Explanation:

The formula for power is given by:

$P= \frac{W}{t}$

Breaking this formula down to its constituents we can determine the units:

$P = \frac{m*a*d}{t}= \frac{(kg)(\frac{m}{s^2})(m)}{s}= \frac{N\cdot m}{s}$

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Example Question #11 : Power

A car of mass 2018 kilograms goes from $0 \frac{m}{s}$ to $50\frac{m}{s}$ in 11.2 seconds .

Assuming perfect traction and ignoring air resistance, estimate the power of the car.

Possible Answers:

$3.6\times10^5 watts$

$1.3\times10^5 watts$

$2.25\times10^5 watts$

$1.19\times10^5 watts$

None of these

Correct answer:

$2.25\times10^5 watts$

Explanation:

First, we need to find the change in energy.

E_f-E_i=W

Our initial energy was zero.

E_f=W

Our final energy consists solely of the final kinetic energy

$.5m{v_f}^2=W$

We plug in our values

$.5\cdot2018\cdot50^2=W$

$2.5\times10^6=W$

Our definition of power is

$P=\frac{W}{\Delta t}$

Plugging in our values again, we get

$P=\frac{2.5\times10^6}{11.2}$

$2.2\times10^5 watts=P$

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Example Question #12 : Power

You are on the ground floor of a parking garage building. You then run up the stairs to the roof, taking 60 seconds to reach a height of 20 meters . You have a mass of 55 kg . Determine your power in watts.

Possible Answers:

None of these

$1.7\times10^4 Watts$

$3.98\times10^4 Watts$

$2.27\times10^4 Watts$

$6.69\times10^4 Watts$

Correct answer:

$1.7\times10^4 Watts$

Explanation:

First, we will need to calculate the total amount of energy gained, also known as the work done.

$W=E_{final}-E_{intial}$

Since we are on the ground floor, we will assume our initial energy to be zero.

Using our formula for gravitational potential energy

$PE_{gravitational}=mgh$

$PE_{gravitational}=55\cdot 9.8\cdot20$

$1.0\times10^6 J=PE_{gravitational}$

These $1.0\times10^6 J$ were obatined in 60 seconds .

Using our definition of power

$P=\frac{W}{\Delta t}$

$P=\frac{1.0\times10^6}{60}$

$1.7\times10^4 Watts=P$

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Example Question #13 : Power

A car of mass 1110kg is accelerated from $10 \frac{m}{s}$ to $35 \frac{m}{s}$ in 2s.

Determine the power output of the car during this time frame.

Possible Answers:

P=3.1*10^5W

P=9.1*10^5W

P=3.7*10^5W

P=4.3*10^5W

None of these

Correct answer:

P=3.1*10^5W

Explanation:

Use the definition of work:

W=E_f-E_i

W=.5mv_f^2-.5mv_i^2

W=.5*1110*35^2-.5*1110*10^2

W=6.2*10^5J

Use the definition of power:

$P=\frac{W}{\Delta t}$

$P=\frac{6.2*10^5}{2}$

P=3.1*10^5W

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Example Question #14 : Power

A person of mass 95kg is riding a ferris wheel of radius 10m . The wheel is spinning at a constant angular velocity of 1rpm . Determine the average power experienced by the rider during their ascent.

Possible Answers:

620W

190W

510W

750W

16W

Correct answer:

620W

Explanation:

If the radius is 10m , then the height gained during the ascent is 20m .

Thus, the gravitational potential energy gained is:

$mgh=PE_{grav}$

$95*9.8*20=PE_{grav}$

$18.6kJ=PE_{grav}$

The ascent took $\frac{1}{2}$ of a rotation, thus:

$t=\frac{1 minute}{rotation}*\frac{60 seconds}{1 minute}*.5rotations$

$t=\30 seconds$

Use the definition of power:

$P=\frac{\Delta E}{\Delta t}$

$P=\frac{18.6kJ}{30 seconds}$

P=620W

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Example Question #15 : Power

A spaceship of mass 7500kg accelerates from $0\frac{m}{s}$ to $\15\frac{m}{s}$ in 2.5s . Determine the average power of the engine during this timeframe.

Possible Answers:

9.9*10^5 W

2.8*10^5 W

3.4*10^5 W

7.7 * 10^5W

1.1*10^5W

Correct answer:

3.4*10^5 W

Explanation:

Calculate work:

E_f-E_i=W

Only kinetic energy involved:

KE_f-KE_i=W

.5mv_f^2-.5mv_i^2=W

Plug in values:

.5*7500*15^2-.5*7500*0^2=W

8.4*10^5 J=W

Definition of power:

$Power=\frac{Work}{time}$

Plug in values:

$Power=\frac{8.4*10^5}{2.5s}$

3.4*10^5 W=Power

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Example Question #16 : Power

A car of mass 1750kg goes from to $100\frac{km}{hr}$ in 7 s . Estimate the power of the engine.

Possible Answers:

1.3*10^4 W

7.7*10^4 W

4.8*10^4W

8.6*10^4 W

10.1*10^4 W

Correct answer:

7.7*10^4 W

Explanation:

Power is defined as:

$Power=\frac{Work}{time}$

Calculate work, with only kinetic energy involved:

E_f-E_i=W

KE_f-KE_i=W

.5mv_f^2-.5mv_i^2=W

Convert $\frac{km}{hr}$ to $\frac{m}{s}$ :

$\frac{100km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=27.8\frac{km}{hr}$

$\frac{0km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=0\frac{km}{hr}$

Plug in values:

.5*1750*27.8^2-.5*1750*0^2=W

5.4*10^5 J=W

Plugging in values:

$Power=\frac{5.4*10^5}{7s}$

7.7*10^4W=Power

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Example Question #11 : Power

A 1500kg car goes from $0\frac{m}{s}$ to $45\frac{m}{s}$ in 4.5s . Estimate how long it would take the car to accelerate to $60\frac{m}{s}$ , assuming the same engine power.

Possible Answers:

10s

Correct answer:

Explanation:

Definition of power:

$J=\frac{W}{\Delta t}$

Definition of work:

W=E_f-E_i

Since all energy in this case is kinetic:

W=.5mv_f^2-.5mv_i^2

Initial velocity is zero:

W=.5mv_f^2

Combine equations:

$J=\frac{.5mv_f^2}{\Delta t}$

Plugg in values for both final velocities and solve for the change in time:

$J=\frac{.5*1500*45^2}{4.5}=\frac{.5*1500*60^2}{\Delta t}$

$\Delta t = 8s$

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Example Question #211 : Work, Energy, And Power

A 1500kg car goes from $0\frac{m}{s}$ to $45\frac{m}{s}$ in 4.5s . Estimate how long it would take the same engine to accelerate a 2500kg truck to the same velocity, assuming the same engine power.

Possible Answers:

7.5s

14s

10.5s

4.5s

20s

Correct answer:

7.5s

Explanation:

Definition of power:

$J=\frac{W}{\Delta t}$

Definition of work:

W=E_f-E_i

Since all energy in this case is kinetic:

W=.5mv_f^2-.5mv_i^2

Initial velocity is zero:

W=.5mv_f^2

Combine equations:

$J=\frac{.5mv_f^2}{\Delta t}$

Plug in values for both vehicles and solve for the change in time:

$J=\frac{.5*1500*45^2}{4.5}=\frac{.5*2500*45^2}{\Delta t}$

$\Delta t = 7.5s$

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Example Question #11 : Power

An engine designed to deliver constant power accelerates a cart from rest. By which of the following factors does velocity change with respect to time?

Possible Answers:

None of these

Exponentially

Linearly

Logarithmic

Square root

Correct answer:

Square root

Explanation:

Definition of power:

$J=\frac{W}{\Delta t}$

Definition of work:

W=E_f-E_i

Since all energy in this case is kinetic:

W=.5mv_f^2-.5mv_i^2

Initial velocity is zero:

W=.5mv_f^2

Combine equations:

$J=\frac{.5mv_f^2}{\Delta t}$

Solving for v_f

$\sqrt{\frac{J \Delta t}{m}}=v_f$

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AP Physics 1 : Power

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #11 : Power

Example Question #11 : Power

Example Question #12 : Power

Example Question #13 : Power

Example Question #14 : Power

Example Question #15 : Power

Example Question #16 : Power

Example Question #11 : Power

Example Question #211 : Work, Energy, And Power

Example Question #11 : Power

All AP Physics 1 Resources

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