University

Physics 1

Physics 1

Fall 2024

In the following scenarios, a force pushes on an object. Determine if the work done by the force is positive, negative, or zero.

- A box is pushed toward the left, and slides along the floor at a constant speed toward the left.
- The same box as in (a). Consider the work done by gravity and work done by the normal force.
- A car turns a corner on a level road at constant speed. Consider the work done by static friction, which causes the centripetal acceleration.
- A crane lowers a mass at constant speed. Consider the work by the tension in the cable.
- Force \(\vec F\) and displacement \(\Delta \vec r\), as shown below.

- Work done by pushing force is positive since force and displacement are in the same (leftward) direction.
- Work done by gravity and the normal force are zero, because these forces are perpendicular to the motion.
- Work done by static friction is zero because \(\vec f\) is always perpendicular to \(\vec v\).
- Work done by the tension is negative because it points up while the motion is down.
- Work done is postive because \(\vec F \cdot \Delta \vec r > 0\).

For the following scenarios, describe the energy transformation taking place. Use \(K\), \(U\), and \(E_{th}\) to refer to kinetic, potential and thermal energy, respectively.

- A diver falls downward, before hitting the water.
- The diver hits the water and rapidly slows.
- A skier is sliding down a slope at a constant speed.
- A box slides up a gentle but slightly rough incline until stopping at the top.

- \(U \rightarrow K\)
- \(K \rightarrow E_{th}\)
- \(U \rightarrow E_{th}\)
- \(K \rightarrow U\) and \(E_{th}\)

*Reasoning*

- As the diver falls, they pick up speed. Elevation loss means \(U\) decreases. Speed increase means \(K\) increases.
- The speed decrease means \(K\) decreases. There is no substantial elevation change, so the energy has been lost to thermal energy in the water.
- The skier’s speed is constant, so \(K\) does not change. There is a loss of elevation, meaning \(U\) is lost. If the skier does not speed up, it must be a friction force that does work. Meaning thermal energy is generated.
- Losing speed means \(K\) is lost. Gaining elevation means \(U\) increases. The “rough” surface implies that friction does work, so thermal energy is also created.

Object \(A\) has mass \(M\), and \(B\) has mass \(2M\). Both are at rest.

Compare their kinetic energies (\(K_A\) and \(K_B\)) after

- They are each pushed with the same force for the same distance \(d\).
- They are each pushed with the same force for the same time \(t\).

- \(K_A = K_B\)
- \(K_A \gt K_B\)

In each case, \(K\) depends on the work done, which is force times distance.

- Work done is the same, so \(\Delta K\) is the same.
- Over time \(t\), \(A\) will accelerate more and move a greater distance. So work done on \(A\) is greater.

A roller coaster rolls from rest down a frictionless track, reaching speed \(v\) at the bottom. If you want the car to go twice as fast at the bottom, by what factor must you increase the height of the track?

The track must be \(4\) times higher.

Energy conservation gives \(mgh = \frac{1}{2} m v^2\), so \(h = \frac{v^2}{2g}\).

So if \(v\) is doubled, the new height will be \(4h\).

Rank the elastic potential enegies stored in the following springs.

\(U_1 = \frac{1}{2} k d^2\)

\(U_2 = \frac{1}{2} k d^2 = U_1\)

\(U_3 = \frac{1}{2} (2k) d^2 = 2 U_1\)

\(U_4 = \frac{1}{2} k (2d)^2 = 4 U_1\)

So

\(U_4 > U_3 > U_2 = U_1\)

A spring-loaded gun shoots out a plastic ball at speed \(v_0\). The spring is then compressed twice the distance it was on the first shot.

- By what factor is the spring’s potential energy increased?
- By what factor is the ball’s launch speed increased?

- \(U = \frac{1}{2} k x^2\), so double \(x\) means four times more energy.
- \(U \rightarrow \frac{1}{2} m v^2\), so four times more energy means double \(v\).

Below are a set of axes on which you are going to draw a potential-energy curve. By doing experiments, you find the following information:

- A particle with energy \(E_1\) oscillates between positions D and E.
- A particle with energy \(E_2\) oscillates between positions C and F.
- A particle with energy \(E_3\) oscillates between positions B and G.
- A particle with energy \(E_4\) enters from the right, turns around at A, then never returns.

Draw a potential-energy curve that is consistent with this information.

The graph shows the potential energy curve of a particle moving along the \(x\)-axis under the influence of a conservative force.

- In which intervals of \(x\) is the force on the particle to the right?
- In which intervals of \(x\) is the force on the particle to the left?
- At what values of \(x\) is the magnitude of the force maximum?
- What values of \(x\) are positions of stable equilibrium?
- What values of \(x\) are positions of unstable equilibrium?
- If the particle is released from rest at \(x=0\) m, will it reach \(x=10\) m?

- \(F_x= - (dU)/dx\) (negative of slope). So \(\vec F_x\) is to the right from \(0\) m to \(2\) m, and from \(5\) m to \(8\) m.
- \(\vec F_x\) is to the left from \(2\) m to \(5\) m, and from \(8\) m to \(10\) m.
- Slope is steepest at \(x=0\) m, \(x=3.5\) m, \(x=6.5\) m, and \(x=10\) m.
- At the minima, \(x=2\) m and \(x=8\) m.
- At the maxima, \(x=5\) m.
- No. The particle will move right until it turns around at about \(x=4\) m.

Two projectiles (with the same mass) are launched over flat ground with the same intial speed. Projectile \(A\) is launched at angle \(10\)°, and projectile \(B\) at angle \(50\)° (above horizontal).

You may ignore air resistance.

- Which projectile has greater speed at its peak height?
- Which projectile has greater speed upon hitting the ground?

Note the equal initial speed means that they have the same total energy at all times throughout their flights: \(K+U =\) constant.

- \(A\) is moving faster. \(K_A \gt K_B\), since \(U_A \lt U_B\) at their peak height.
- Same speed. \(K_A = K_B\), since \(U_A=U_B=0\) at the ground.

Consider a particle on which several forces act, one of which is known to be constant in time: \(\vec{F}_1 = 3 \hat{\imath} + 4 \hat{\jmath}\) N

As a result, the particle moves along a straight path from a Cartesian coordinate of (\(-1\) m, \(2\) m) to (\(5\) m, \(6\) m). What is the work done by \(\vec F_1\) ?

Since the force is constant and the path is straight, \(W = \vec F \cdot \Delta \vec s\).

\(W= (3 \hat{\imath} + 4 \hat{\jmath}) \cdot (6\hat{\imath} + 4 \hat{\jmath}) = 18+16 =34\) N·m

An electron in a television tube is accelerated uniformly from rest to a speed of \(8.4 \times 10^7\) m/s over a distance of \(2.5\) cm. What is the power delivered to the electron at the instant that its displacement is \(1.0\) cm?

The electron has a mass of \(9.11 \times 10^{-31}\) kg.

Work done is the gain in kinetic energy: \(W=Fd= \Delta K =\frac{1}{2} m v^2\).

So with \(d=2.5\) cm, you can find the force on the electron. From this, acceleration is

\(a = F/m = 1.41 \times 10^{17}\) m/s²

Now with \(d= 1.0\) cm, find velocity with \(v^2 = 2 a \Delta s\).

\(v = 5.31 \times 10^{7}\) m/s

Then \(P = F \cdot v\).

\(P = 6.83\) μW

A \(40\) kg crate is pushed uphill at constant velocity a distance \(8.0\) m along a \(30\)° incline by the horizontal force \(\vec F\). The coefficient of kinetic friction between the crate and the incline is \(\mu_k\) = \(0.40\).

Calculate the work done by

- the applied force
- the normal force
- the frictional force
- the gravitational force
- the net force

- \(3.46\) kJ
- \(0\)
- \(–1.89\) kJ
- \(–1.57\) kJ
- \(0\)

Consider the free-body diagram:

Note that the normal force is perpendicular to displacement, and therefore does zero work.

\(\vec{F}=m\vec{a}\) with the axes shown gives the following:

\(\Sigma F_x = 0 = F - N \sin \theta - \mu
N \cos \theta\)

\(\Sigma F_y = 0 = N \cos \theta - \mu N \sin
\theta - mg\)

where \(\theta = 30^{\circ}\).

The second equation yeilds \(N = \frac{mg}{\cos \theta - \mu \sin \theta} = 588.57\) N.

With this, the first equation yeilds \(F = 498.17\) N.

With these you can find work done with \(W = \vec F \cdot \Delta \vec s\).

Since \(\vec{a} = 0\), \(\vec{F}_{\textup{net}} = 0\). So the net force does zero work.

A 10-kg box slides along a track at 8 m/s. As the box slides up a ramp, it gains \(1.0\) m of elevation, while friction does \(-100\) J of work on it. How fast is it moving afterward?

The change in energy \(\Delta (K+U)\) equals work done by friction (which is negative because energy is lost).

\(K_f + U_f - (K_i +U_i) = W\)

\(K_f = \frac{1}{2} m v_f^2\)

\(U_f = mgh\)

\(K_i = \frac{1}{2} m v_i^2\)

\(U_i = 0\)

Plug these in and solve for \(v_f\).

\(v_f^2 = v_i^2 - 2gh + 2\frac{W}{m} = 8^2 - 2(9.8)(1.0) - 2(100)/10 = 24.4\) m²/s²

\(v_f = 4.9\) m/s

*Last modified: Tue October 22 2024, 09:25 PM.*