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AP Physics 1 : Newtonian Mechanics

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Example Questions

Example Question #31 : Work, Energy, And Power

A rocket of mass 1560kg $\displaystyle 1560kg$ is motionless at the origin. The rocket then fires. At location <500,500>m $\displaystyle < 500,500>m$, it is traveling at $<250,250>\frac{m}{s}$ $\displaystyle < 250,250>\frac{m}{s}$.

Determine the magnitude of the average force of the rocket.

Possible Answers:

$\displaystyle 4.56*10^5N$

$\displaystyle 9.34*10^4N$

$\displaystyle 1.37*10^5N$

None of these

$\displaystyle 7.96*10^5N$

Correct answer:

$\displaystyle 1.37*10^5N$

Explanation:

Determining magnitude of final velocity:

$|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}$ $\displaystyle |\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}$

Plugging in values:

$|\overrightarrow{v}|=\sqrt{250^2+250^2}$ $\displaystyle |\overrightarrow{v}|=\sqrt{250^2+250^2}$

$|\overrightarrow{v}|=353.6\frac{m}{s}$ $\displaystyle |\overrightarrow{v}|=353.6\frac{m}{s}$

Using definition of kinetic energy:

$\displaystyle KE=.5mv^2$

Plugging in values:

$\displaystyle KE=.5*1560*353.6^2$

$\displaystyle KE=9.75*10^7J$

Since there was no initial kinetic energy:

$\displaystyle W=9.75*10^7J$

Finding total distance:

$|\overrightarrow{d}|=\sqrt{d_x^2+d_y^2}$ $\displaystyle |\overrightarrow{d}|=\sqrt{d_x^2+d_y^2}$

$|\overrightarrow{d}|=\sqrt{500^2+500^2}$ $\displaystyle |\overrightarrow{d}|=\sqrt{500^2+500^2}$

d=707m $\displaystyle d=707m$

Using

W=F*d $\displaystyle W=F*d$

Plugging in values:

$\displaystyle F*707=9.75*10^7$

$\displaystyle F=1.37*10^5N$

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Example Question #31 : Newtonian Mechanics

A ball of mass 30g $\displaystyle 30g$ is thrown at a target. The ball strikes with a velocity of $20\frac{m}{s}$ $\displaystyle 20\frac{m}{s}$ and bounces back with equal magnitude. Determine the work done on the ball.

Possible Answers:

$\displaystyle 6J$

None of these

12J $\displaystyle 12J$

$\displaystyle 0J$

-12J $\displaystyle -12J$

Correct answer:

$\displaystyle 0J$

Explanation:

There is no net work done on the ball. The wall stopped the ball, doing negative work, then accelerated the ball, doing positive work. These end up canceling each other out.

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Example Question #33 : Work, Energy, And Power

During time period $\displaystyle A$, a rocket ship deep in space of mass $2.55*10^5\textup{ kg}$ $\displaystyle 2.55*10^5\textup{ kg}$ travels from $<0,10>\textup{ km}$ $\displaystyle < 0,10>\textup{ km}$ to $<5,5>\textup{ km}$ $\displaystyle < 5,5>\textup{ km}$. During time period $\displaystyle B$, the rocket fires. During time period $\displaystyle C$, the rocket travels from $<-15,0>\textup{ km}$ $\displaystyle < -15,0>\textup{ km}$ to $<-30,-5>\textup{ km}$ $\displaystyle < -30,-5>\textup{ km}$.

Time periods $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ took $10\textup{ s}$ $\displaystyle 10\textup{ s}$ each.

Determine the work done during time period $\displaystyle B$.

Possible Answers:

$2.15*10^{11}\textup{ J}$ $\displaystyle 2.15*10^{11}\textup{ J}$

None of these

$8.44*10^{11}\textup{ J}$ $\displaystyle 8.44*10^{11}\textup{ J}$

$2.55*10^{11} \textup{ J}$ $\displaystyle 2.55*10^{11} \textup{ J}$

$3.95*10^{11}\textup{ J}$ $\displaystyle 3.95*10^{11}\textup{ J}$

Correct answer:

$2.55*10^{11} \textup{ J}$ $\displaystyle 2.55*10^{11} \textup{ J}$

Explanation:

Using

$\displaystyle W=KE_f-KE_i$

Determining initial kinetic energy:

$\displaystyle KE=.5mv^2$

$v=\frac{d}{t}$ $\displaystyle v=\frac{d}{t}$

$d=\sqrt((x_f-x_i)^2+(y_f-y_i)^2)$ $\displaystyle d=\sqrt((x_f-x_i)^2+(y_f-y_i)^2)$

Combining equations

$KE=.5m(\frac{\sqrt((x_f-x_i)^2+(y_f-y_i)^2)}{t})^2$ $\displaystyle KE=.5m(\frac{\sqrt((x_f-x_i)^2+(y_f-y_i)^2)}{t})^2$

Converting $\displaystyle km$ to $\displaystyle m$ and plugging in values:

$KE=.5*2.55*10^5(\frac{\sqrt((5000-0)^2+(5000-10000)^2)}{10})^2$ $\displaystyle KE=.5*2.55*10^5(\frac{\sqrt((5000-0)^2+(5000-10000)^2)}{10})^2$

$KE_i=6.375*10^{10}J$ $\displaystyle KE_i=6.375*10^{10}J$

Determining final kinetic energy:

$\displaystyle KE=.5mv^2$

$v=\frac{d}{t}$ $\displaystyle v=\frac{d}{t}$

$d=\sqrt((x_f-x_i)^2+(y_f-y_i)^2)$ $\displaystyle d=\sqrt((x_f-x_i)^2+(y_f-y_i)^2)$

Combining equations

$KE=.5m(\frac{\sqrt((x_f-x_i)^2+(y_f-y_i)^2)}{t})^2$ $\displaystyle KE=.5m(\frac{\sqrt((x_f-x_i)^2+(y_f-y_i)^2)}{t})^2$

Converting $\displaystyle km$ to $\displaystyle m$ and plugging in values:

$KE=.5*2.55*10^5(\frac{\sqrt((-15000+30000)^2+(0+5000)^2)}{10})^2$ $\displaystyle KE=.5*2.55*10^5(\frac{\sqrt((-15000+30000)^2+(0+5000)^2)}{10})^2$

$KE_f=3.188*10^{11}J$ $\displaystyle KE_f=3.188*10^{11}J$

Plugging in values:

$W=3.188*10^{11}-6.375*10^{10}$ $\displaystyle W=3.188*10^{11}-6.375*10^{10}$

$W=2.55*10^{11}J$ $\displaystyle W=2.55*10^{11}J$

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Example Question #32 : Work, Energy, And Power

The slingshot is a popular attraction at many amusement parks. During the ride, riders sit in a ball and are propelled vertically by a massive slingshot. A certain sling shot uses 25 springs, each with a constant of $1200 \frac{N}{m}$ $\displaystyle 1200 \frac{N}{m}$. If the springs are stretched 1.5m $\displaystyle 1.5m$ and the ball reaches a maximum height of 60 m $\displaystyle 60 m$ when no riders are in it, what is the total mass of the ball?

$g = 10\frac{m}{s^2}$ $\displaystyle g = 10\frac{m}{s^2}$

Possible Answers:

115.1kg $\displaystyle 115.1kg$

129.7kg $\displaystyle 129.7kg$

155.5kg $\displaystyle 155.5kg$

56.25kg $\displaystyle 56.25kg$

95.2kg $\displaystyle 95.2kg$

Correct answer:

56.25kg $\displaystyle 56.25kg$

Explanation:

We can use the expression for conservation of energy to solve this problem:

$\displaystyle E = U_i+K_i = U_f+K_f$

Given the problem statement, we know that the ball has no velocity at the initial and final states, so we can remove kinetic energy from the equation:

$\displaystyle U_i = U_f$

The initial potential energy is stored in the springs and the final potential energy is gravitational, so we can write:

$25(\frac{1}{2}kx^2) = mgh$ $\displaystyle 25(\frac{1}{2}kx^2) = mgh$

We multiply by 25 because there are 25 springs.

Rearranging for mass, we get:

$m=\frac{25kx^2}{2gh}$ $\displaystyle m=\frac{25kx^2}{2gh}$

We know all of these values, allowing us to solve:

$m = \frac{25(1200\frac{N}{m})(1.5)^2}{2(10\frac{m}{s^2})(60m)} =56.25kg$ $\displaystyle m = \frac{25(1200\frac{N}{m})(1.5)^2}{2(10\frac{m}{s^2})(60m)} =56.25kg$

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Example Question #73 : Ap Physics 1

Suppose that you're an engineer, and you have been asked to develop a ramp that makes it easier to lift things up to a $2.5\: m$ $\displaystyle 2.5\: m$ high platform. If the ramp is $5\: m$ $\displaystyle 5\: m$ long, how does the force used to lift the object change?

Possible Answers:

The force used to lift the object via the ramp is cut by one fourth

The force used to lift the object via the ramp is doubled

The force used to lift the object via the ramp is halved

The force used to lift the object via the ramp is quadrupled

The force used to lift the object via the ramp is the same

Correct answer:

The force used to lift the object via the ramp is halved

Explanation:

For this question, we're told that an object is normally lifted a certain vertical distance. However, a ramp is to be added in order to make it easier to lift the object to the desired position, and we're asked to find out how much easier it will be by determining how the force required to lift it will change.

First, we must approach this problem from the perspective of energy. Specifically, we need to look at the change in gravitational potential energy. When the object is lifted a certain distance, its gravitational potential energy will increase according to the following expression:

$\Delta U=mgh$ $\displaystyle \Delta U=mgh$

Furthermore, it's important to realize that the change in mechanical potential energy only cares about the final and initial positions; it does NOT care about the path taken to get from initial to final. Therefore, the change in mechanical potential energy for lifting the object directly up is exactly the same as if the object were to be moved up a ramp to the same vertical location.

What's more is that we can realize the gravitational potential energy will be increasing as it is lifted, thus we need to put energy into this process by doing work. We can write the expression for work as follows:

W=Fd $\displaystyle W=Fd$

Since the amount of work done on both processes is the same, we can set the two expressions equal to each other as follows:

$F_{lift}d_{lift}=F_{ramp}d_{ramp}$ $\displaystyle F_{lift}d_{lift}=F_{ramp}d_{ramp}$

By rearranging, we obtain:

$\frac{F_{ramp}}{F_{lift}}=\frac{d_{lift}}{d_{ramp}}$ $\displaystyle \frac{F_{ramp}}{F_{lift}}=\frac{d_{lift}}{d_{ramp}}$

The expression shown above tells us how the force needed to transfer the object via the ramp is different from the force needed to lift the object directly.

$\frac{F_{ramp}}{F_{lift}}=\frac{2.5\: m}{5\: m}=0.5$ $\displaystyle \frac{F_{ramp}}{F_{lift}}=\frac{2.5\: m}{5\: m}=0.5$

Thus, the force needed to move the object via the ramp is halved compared to the original force. Essentially, since we doubled the distance, we halved the force.

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Example Question #36 : Work, Energy, And Power

A weightlifter doing a bench press lifts a $102\ kg$ $\displaystyle 102\ kg$ weight from the base of his chest to a height that is $0.7\ m$ $\displaystyle 0.7\ m$ above his chest. How much work has he done after a single lift?

Possible Answers:

$700\ J$ $\displaystyle 700\ J$

$625\ J$ $\displaystyle 625\ J$

$955\ J$ $\displaystyle 955\ J$

$815\ J$ $\displaystyle 815\ J$

$1100\ J$ $\displaystyle 1100\ J$

Correct answer:

$700\ J$ $\displaystyle 700\ J$

Explanation:

Work is equal to force multiplied by the distance traveled in the direction that the force is applied. The force in this problem is equal to the gravitational force that is created by the weight because that is the minimum force that the weightlifter must apply in order to move the weight. The distance is the distance from the weightlifters chest to the peak of his lift.

W=Fd $\displaystyle W=Fd$

$\displaystyle W=102*9.81*0.7$

W=700J $\displaystyle W=700J$

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Example Question #72 : Ap Physics 1

A quarterback throws a football a horizontal distance of 50 yd $\displaystyle 50 yd$ to a wide receiver. The ball was airborne for 5 s $\displaystyle 5 s$. The ball had an initial speed of 50 mph $\displaystyle 50 mph$. The ball has a mass of .43kg $\displaystyle .43kg$.

$1mph=.447\frac{m}{s}$ $\displaystyle 1mph=.447\frac{m}{s}$

$\displaystyle 1 yd=.914 m$

Estimate the distance that the quarterback's hand was in contact with the ball during the throw, given that the throw took .35 s $\displaystyle .35 s$.

Possible Answers:

2.75m $\displaystyle 2.75m$

1.25m $\displaystyle 1.25m$

3.91m $\displaystyle 3.91m$

8.64m $\displaystyle 8.64m$

None of these

Correct answer:

3.91m $\displaystyle 3.91m$

Explanation:

Using

$\displaystyle W=E_f-E_i$

W=F*d $\displaystyle W=F*d$

All energy will be kinetic:

$\displaystyle W=.5mv_f^2-.5mv_i^2$

The initial velocity will be zero as the ball is motionless

$\displaystyle W=.5mv_f^2$

Combining equations:

$\displaystyle F*d=.5mv_f^2$

Using:

$F*\Delta t=p_f-p_i$ $\displaystyle F*\Delta t=p_f-p_i$

p=mv $\displaystyle p=mv$

Combining equations:

$F*\Delta t=mv_f-mv_i$ $\displaystyle F*\Delta t=mv_f-mv_i$

The initial velocity is zero

$F*\Delta t=mv_f$ $\displaystyle F*\Delta t=mv_f$

Combining equations and solving for $\displaystyle d$

$d=.5\Delta t *v_f$ $\displaystyle d=.5\Delta t *v_f$

Converting mph $\displaystyle mph$ to $\frac{m}{s}$ $\displaystyle \frac{m}{s}$

$50mph*\frac{.447 m/s}{1 mph}=22.35\frac{m}{s}$ $\displaystyle 50mph*\frac{.447 m/s}{1 mph}=22.35\frac{m}{s}$

Plugging in values:

$\displaystyle d=.5*.35 *22.35$

$\displaystyle d=3.91meters$

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Example Question #32 : Work

$\displaystyle 1 mile=1.61km$

A 2300kg $\displaystyle 2300kg$ funny car goes from rest to 332mph $\displaystyle 332mph$ in 3.5s $\displaystyle 3.5s$. Determine the average force exerted on the car.

Possible Answers:

125kN $\displaystyle 125kN$

184kN $\displaystyle 184kN$

None of these

34kN $\displaystyle 34kN$

104kN $\displaystyle 104kN$

Correct answer:

34kN $\displaystyle 34kN$

Explanation:

Converting mph $\displaystyle mph$ to $\frac{m}{s}$ $\displaystyle \frac{m}{s}$

$\frac{332miles}{hour}*\frac{1.61km}{1 mile}*\frac{1000m}{1km}*\frac{1 hour}{3600s}=148\frac{m}{s}$ $\displaystyle \frac{332miles}{hour}*\frac{1.61km}{1 mile}*\frac{1000m}{1km}*\frac{1 hour}{3600s}=148\frac{m}{s}$

F=ma $\displaystyle F=ma$

$F=m\frac{\Delta v}{\Delta t}$ $\displaystyle F=m\frac{\Delta v}{\Delta t}$

$F=2300kg\frac{148-0}{10}$ $\displaystyle F=2300kg\frac{148-0}{10}$

F=34kN $\displaystyle F=34kN$

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Example Question #74 : Ap Physics 1

A 2300kg $\displaystyle 2300kg$ funny car goes from rest to 332mph $\displaystyle 332mph$ in 3.5s $\displaystyle 3.5s$. Determine the average force exerted on the car, and use that to determine how far the car traveled in that time period.

Possible Answers:

None of these

899m $\displaystyle 899m$

741m $\displaystyle 741m$

421m $\displaystyle 421m$

179m $\displaystyle 179m$

Correct answer:

741m $\displaystyle 741m$

Explanation:

Converting mph $\displaystyle mph$ to $\frac{m}{s}$ $\displaystyle \frac{m}{s}$

F=ma $\displaystyle F=ma$

$F=m\frac{\Delta v}{\Delta t}$ $\displaystyle F=m\frac{\Delta v}{\Delta t}$

$F=2300kg\frac{148-0}{10}$ $\displaystyle F=2300kg\frac{148-0}{10}$

F=34kN $\displaystyle F=34kN$

Since the initial velocity is zero, initial kinetic energy is zero

$\displaystyle W=KE_f-KE_i$

W=KE_f $\displaystyle W=KE_f$

$\displaystyle W=.5mv_f^2$

Plugging in values:

$\displaystyle W=.5*2300*148^2$

$\displaystyle W=2.52*10^7J$

W=F*d $\displaystyle W=F*d$

$\displaystyle 2.52*10^7J=34000*d$

d=741m $\displaystyle d=741m$

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Example Question #39 : Work, Energy, And Power

A spaceship of mass 7500kg $\displaystyle 7500kg$ is motionless in space. The rocket is turned on and provides a constant force of 1000N $\displaystyle 1000N$. Assume the mass of mass due to spent fuel is negligible.

Determine the work done by the rocket in 10s. $\displaystyle 10s.$

Possible Answers:

$6633N\cdot m$ $\displaystyle 6633N\cdot m$

$1138N\cdot m$ $\displaystyle 1138N\cdot m$

None of these

$1133N\cdot m$ $\displaystyle 1133N\cdot m$

$-6633N\cdot m$ $\displaystyle -6633N\cdot m$

Correct answer:

$6633N\cdot m$ $\displaystyle 6633N\cdot m$

Explanation:

F=ma $\displaystyle F=ma$

Plugging in values

$\displaystyle 1000=7500*a$

$a=.133\frac{m}{s}$ $\displaystyle a=.133\frac{m}{s}$

Using

$\displaystyle v_i+a*t=v_f$

$\displaystyle 0+.133*10=v_f$

$v_f=1.33\frac{m}{s}$ $\displaystyle v_f=1.33\frac{m}{s}$

Using

$\displaystyle KE=.5mv^2$

$\displaystyle KE=.5*7500*1.33^2$

$\displaystyle KE=6633J$

Using

$\displaystyle W=KE_f-KE_i$

Initially the space ship is motionless, so it has $\displaystyle 0$ kinetic energy

$\displaystyle W=6633-0$

$W=6633N\cdot m$ $\displaystyle W=6633N\cdot m$

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AP Physics 1 : Newtonian Mechanics

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #31 : Work, Energy, And Power

Example Question #31 : Newtonian Mechanics

Example Question #33 : Work, Energy, And Power

Example Question #32 : Work, Energy, And Power

Example Question #73 : Ap Physics 1

Example Question #36 : Work, Energy, And Power

Example Question #72 : Ap Physics 1

Example Question #32 : Work

Example Question #74 : Ap Physics 1

Example Question #39 : Work, Energy, And Power

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