- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
6.1 Work Done by a Constant Force
$
W=F s
$
\(1 \mathrm{~N} \cdot \mathrm{m}=1\) joule \((\mathrm{J})\)
\(W=(F \cos \theta) s\)
\(\begin{aligned} & \cos 0^{\circ}=1 \\ & \cos 90^{\circ}=0 \\ & \cos 180^{\circ}=-1\end{aligned}\)
Example 1 Pulling a Suitcase-on-Wheels
Find the work done if the force is \(45.0-\mathrm{N}\), the angle is 50.0 degrees, and the displacement is \(75.0 \mathrm{~m}\).
$
\begin{aligned}
& W=(F \cos \theta) S=\left[(45.0 \mathrm{~N}) \cos 50.0^{\circ}\right](75.0 \mathrm{~m}) \\
& =2170 \mathrm{~J}
\end{aligned}
$
\(\begin{gathered}W=(F \cos 0) s=F s \\ W=(F \cos 180) s=-F s\end{gathered}\)
Example 3 Accelerating a Crate
The truck is accelerating at a rate of \(+1.50 \mathrm{~m} / \mathrm{s}^2\). The mass of the crate is \(120-\mathrm{kg}\) and it does not slip. The magnitude of the displacement is \(65 \mathrm{~m}\).
What is the total work done on the crate by all of the forces acting on it?
Ans:
The angle between the displacement and the normal force is 90 degrees.
The angle between the displacement and the weight is also 90 degrees.
$
W=(F \cos 90) s=0
$
The angle between the displacement and the friction force is 0 degrees.
$
\begin{aligned}
& f_s=m a=(120 \mathrm{~kg})\left(1.5 \mathrm{~m} / \mathrm{s}^2\right)=180 \mathrm{~N} \\
& W=[(180 \mathrm{~N}) \cos 0](65 \mathrm{~m})=1.2 \times 10^4 \mathrm{~J}
\end{aligned}
$
6.2 The Work-Energy Theorem and Kinetic Energy
Consider a constant net external force acting on an object.
The object is displaced a distance \(s\), in the same direction as the net force.
The work is simply \(W=\left(\sum F\right) s=(m a)_S\)
DEFINITION OF KINETIC ENERGY
The kinetic energy KE of an object with mass \(m\) and speed \(v\) is given by
$
\mathrm{KE}=\frac{1}{2} m v^2
$
THE WORK-ENERGY THEOREM
When a net external force does work on an object, the kinetic energy of the object changes by the amount of work done on it:
$$
W=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{o}}=\frac{1}{2} m v_{\mathrm{f}}^2-\frac{1}{2} m v_o^2
$$
Example 4 Deep Space 1
The mass of the space probe is \(474-\mathrm{kg}\) and its initial velocity is \(275 \mathrm{~m} / \mathrm{s}\). If the \(56.0-\mathrm{mN}\) force acts on the probe through a displacement of \(2.42 \times 10^9 \mathrm{~m}\), what is its final speed?
▶️Answer/Explanation
\(\mathrm{W}=\frac{1}{2} m v_{\mathrm{f}}^2-\frac{1}{2} m v_o^2\)
\(\mathrm{W}=\left[\left(\sum F\right) \cos \theta\right] s\)
\(\begin{gathered}{\left[\left(\sum \mathrm{F}\right) \cos \theta\right]_S=\frac{1}{2} m v_{\mathrm{f}}^2-\frac{1}{2} m v_o^2} \\ \left(5.60 \times 10^{-2} \mathrm{~N}\right) \cos 0^{\circ}\left(2.42 \times 10^9 \mathrm{~m}\right)=\frac{1}{2}(474 \mathrm{~kg}) v_{\mathrm{f}}^2-\frac{1}{2}(474 \mathrm{~kg})(275 \mathrm{~m} / \mathrm{s})^2\end{gathered}\)
\(v_f=805 \mathrm{~m} / \mathrm{s}\)
In this case the net force is \(\sum F=m g \sin 25^{\circ}-f_k\)
Conceptual Example 6 Work and Kinetic Energy
A satellite is moving about the earth in a circular orbit and an elliptical orbit. For these two orbits, determine whether the kinetic energy of the satellite changes during the motion.
6.3 Gravitational Potential Energy
$
\begin{gathered}
W=(F \cos \theta) s \\
W_{\text {gravity }}=m g\left(h_o-h_f\right)
\end{gathered}
$
\(W_{\text {gravity }}=m g\left(h_o-h_f\right)\)
Example 7 A Gymnast on a Trampoline
The gymnast leaves the trampoline at an initial height of \(1.20 \mathrm{~m}\) and reaches a maximum height of \(4.80 \mathrm{~m}\) before falling back down. What was the initial speed of the gymnast?
▶️Answer/Explanation
Ans:
\(\begin{gathered}\mathrm{W}=\frac{1}{2} m v_{\mathrm{f}}^2-\frac{1}{2} m v_o^2 \\ W_{\text {gravity }}=m g\left(h_o-h_f\right)\end{gathered}\)
\(m g\left(h_o-h_f\right)=-\frac{1}{2} m v_o^2\)
\(v_o=\sqrt{-2 g\left(h_o-h_f\right)}\)
\(v_o=\sqrt{-2\left(9.80 \mathrm{~m} / \mathrm{s}^2\right)(1.20 \mathrm{~m}-4.80 \mathrm{~m})}=8.40 \mathrm{~m} / \mathrm{s}\)
$
W_{\text {gravity }}=m g h_o-m g h_f
$
DEFINITION OF GRAVITATIONAL POTENTIAL ENERGY
\The gravitational potential energy PE is the energy that an object of mass \(m\) has by virtue of its position relative to the surface of the earth. That position is measured by the height \(h\) of the object relative to an arbitrary zero level:
$
\mathrm{PE}=m g h
$
\(1 \mathrm{~N} \cdot \mathrm{m}=1\) joule \((\mathrm{J})\)
6.4 Conservative Versus Nonconservative Forces
DEFINITION OF A CONSERVATIVE FORCE
Version 1 A force is conservative when the work it does on a moving object is independent of the path between the object’s initial and final positions.
Version 2 A force is conservative when it does no work on an object moving around a closed path, starting and finishing at the same point.
Table 6.2 Some Conservative and Nonconservative Forces
Conservative Forces
Gravitational force (Ch. 4)
Elastic spring force (Ch. 10)
Electric force (Ch. 18, 19)
Nonconservative Forces
Static and kinetic frictional forces
Air resistance
Tension
Normal force
Propulsion force of a rocket
Version 1 A force is conservative when the work it does on a moving object is independent of the path between the object’s initial and final positions.
▶️Answer/Explanation
\(W_{\text {gravity }}=m g\left(h_o-h_f\right)\)
Version 2 A force is conservative when it does no work on an object moving around a closed path, starting and finishing at the same point.
$
W_{\text {gravity }}=m g\left(h_o-h_f\right) \quad h_o=h_f
$
An example of a nonconservative force is the kinetic frictional force.
$
W=(F \cos \theta) s=f_k \cos 180^{\circ} s=-f_k s
$
The work done by the kinetic frictional force is always negative. Thus, it is impossible for the work it does on an object that moves around a closed path to be zero.
The concept of potential energy is not defined for a nonconservative force.
In normal situations both conservative and nonconservative forces act simultaneously on an object, so the work done by the net external force can be written as
$
W=W_c+W_{n c}
$
\(W=\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{o}}=\Delta \mathrm{KE}\)
\(W_c=W_{\text {gravity }}=m g h_o-m g h_f=\mathrm{PE}_{\mathrm{o}}-\mathrm{PE}_{\mathrm{f}}=-\Delta \mathrm{PE}\)
\(W=W_c+W_{n c}\)
\(\Delta \mathrm{KE}=-\Delta \mathrm{PE}+W_{n c}\)
THE WORK-ENERGY THEOREM
$
W_{n c}=\Delta \mathrm{KE}+\Delta \mathrm{PE}
$
6.5 The Conservation of Mechanical Energy
$
\begin{gathered}
W_{n c}=\Delta \mathrm{KE}+\Delta \mathrm{PE}=\left(\mathrm{KE}_{\mathrm{f}}-\mathrm{KE}_{\mathrm{o}}\right)+\left(\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{\mathrm{o}}\right) \\
W_{n c}=\left(\mathrm{KE}_{\mathrm{f}}+\mathrm{PE}_{\mathrm{f}}\right)+\left(\mathrm{KE}_{\mathrm{o}}+\mathrm{PE}_{\mathrm{o}}\right) \\
W_{n c}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{o}}
\end{gathered}
$
If the net work on an object by nonconservative forces is zero, then its energy does not change:
\(E_f=E_o\)
THE PRINCIPLE OF CONSERVATION OF MECHANICAL ENERGY
The total mechanical energy ( \(E=K E+P E)\) of an object remains constant as the object moves, provided that the net work done by external nononservative forces is zero.
Example 8 A Daredevil Motorcyclist
A motorcyclist is trying to leap across the canyon by driving horizontally off a cliff \(38.0 \mathrm{~m} / \mathrm{s}\). Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.
\(\begin{gathered}\mathrm{E}_{\mathrm{f}}=\mathrm{E}_{\mathrm{o}} \\ m g h_f+\frac{1}{2} m v_f^2=m g h_o+\frac{1}{2} m v_o^2 \\ g h_f+\frac{1}{2} v_f^2=g h_o+\frac{1}{2} v_o^2\end{gathered}\)
\(\begin{gathered}g h_f+\frac{1}{2} v_f^2=g h_o+\frac{1}{2} v_o^2 \\ v_f=\sqrt{2 g\left(h_o-h_f\right)+v_o^2} \\ v_f=\sqrt{2\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)(35.0 \mathrm{~m})+(38.0 \mathrm{~m} / \mathrm{s})^2}=46.2 \mathrm{~m} / \mathrm{s}\end{gathered}\)
Conceptual Example 9 The Favorite Swimming Hole
The person starts from rest, with the rope held in the horizontal position, swings downward, and thel go of the rope. Three force: act on him: his weight, the tension in the rope, and the force of air resistance.
Can the principle of conservation of energy be used to calculate his final speed?
6.6 Nonconservative Forces and the Work-Energy Theorem
THE WORK-ENERGY THEOREM
$
\begin{gathered}
W_{n c}=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{o}} \\
W_{n c}=\left(m g h_f+\frac{1}{2} m v_f^2\right)-\left(m g h_o+\frac{1}{2} m v_o^2\right)
\end{gathered}
$
Example 11 Fireworks
Assuming that the nonconservative force generated by the burning propellant does \(425 \mathrm{~J}\) of work, what is the final speed of the rocket. Ignore air resistance.
▶️Answer/Explanation
$
\begin{aligned}
& W_{n c}=\left(m g h_f+\frac{1}{2} m v_f^2\right)- \\
& \left(m g h_o+\frac{1}{2} m v_o^2\right)
\end{aligned}
$
\(\begin{aligned} & W_{n c}=m g h_f-m g h_o+\frac{1}{2} m v_f^2-\frac{1}{2} m v_o^2 \\ & W_{n c}=m g\left(h_f-h_o\right)+\frac{1}{2} m v_f^2 \\ & 425 \mathrm{~J}=(0.20 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^2\right)(29.0 \mathrm{~m}) \\ & +\frac{1}{2}(0.20 \mathrm{~kg}) v_f^2 \\ & \quad v_f=61 \mathrm{~m} / \mathrm{s}\end{aligned}\)
6.7 Power
DEFINITION OF AVERAGE POWER
Average power is the rate at which work is done, and it is obtained by dividing the work by the time required to perform the work.
$
\bar{P}=\frac{\text { Work }}{\text { Time }}=\frac{W}{t}
$
joule \(/ \mathrm{s}=\mathrm{watt}(\mathrm{W})\)
$
\bar{P}=\frac{\text { Change in energy }}{\text { Time }}
$
1 horsepower \(=550\) foot \(\cdot\) pounds/ second \(=745.7\) watts
$
\bar{P}=F \bar{v}
$
\({ }^{\mathrm{a}}\) For a young \(70-\mathrm{kg}\) male.
6.8 Other Forms of Energy and the Conservation of Energy
THE PRINCIPLE OF CONSERVATION OF ENERGY
Energy can neither be created not destroyed, but can only be converted from one form to another.
6.9 Work Done by a Variable Force
Constant Force
$
W=(F \cos \theta) s
$
Variable Force
$
W \approx(F \cos \theta)_1 \Delta s_1+(F \cos \theta)_2 \Delta s_2+\cdots
$