Physics Bootcamp

Ignoring Drag and Kinetic Friction forces, the forces on the skier are (1) Weight and (2) Normal. Normal force will do no work since, at each instant, it is perpendicular to the displacement. Thus, we have only a conservative force on the skier. Therefore, his energy cannot change.

Equating the energy of the skier at the two instants in the problem statement,

The mass \(m\) cancels out. Solving for \(v_f \) we get

Now, we put in the numbers.

The figure here shows the height \(h\) of the bob from the zero potential energy reference. From the adjacent side of the triangle and the length of the string, we get

\begin{equation*} h = l - l \, \cos\, \theta. \end{equation*}

At initial instant, the energy of the pendulum bob is all potential energy since it is at rest.

(a) Let \(v_a \) stand for the speed when the bob is at the bottom of the swing. At this instant \(h=0\text{.}\) Using the conservation of energy we get the following equation.

Therefore,

(b) Let \(v_b \) stand for the speed when the bob is at the place when the pendulum angle is \(20^\circ\text{.}\) Using the conservation of energy we get the following equation.

where \(E = 0.5257\text{ J}\) and

Solving for \(v_b \text{.}\)

Numerically, we have

(a) Using the formula for the spring potential energy we get

(b) When the block crosses the equilibrium point, \(x = 0\text{.}\) That means potential energy is zero at this point. Let \(v_b\) be the speed when passing through the equilibrium point (see Figure 8.7.5).

When the block was releases at the initial instant in part (a), all the energy was potential energy. Therefore, we get the following equation when we apply energy conservation.

Therefore,

(c) 20 cm from the place the motion began, the block would be 8 cm from the equilibrium, because it will go 12 cm to equilibrium and then stretch another 8 cm. Let's call this \(x_c\) and the speed there as \(v_c\text{.}\) Therefore, the energy conservation will give

Solving for \(v_c \) gives

Now, we put the numbers.

Therefore,

(a) We just need to find the stretching that took place when we hung the block. Let this distance be \(d\text{.}\) Balancing gravitational and spring forces when the block is at O gives

Therefore,

(b) When the block is at the lowest point, it has only potential energy, which is a sum of the spring and gravitational potential energies. The net deformation of the spring is

This will give

The gravitational potential energy will be with respect to O. Since this point is below O, the gravitaional potential energy will be negative.

Therefore, the total energy at this point is

(c) Energy is conserved here. When the block returns to O, it has zero gravitational potential energy, but it still has spring potential energy for the stretch \(d \) found in (a) above.

Applyig energy conservation between this point and the lowest point of motion in part (b) gives the following for speed \(v \) when at O.

This gives

We balance forces along the incline when the block is not yet pushed. The component of weight along the incline is balanced by the spring force from compression by \(2\text{ cm}\text{.}\)

\begin{align*} kx \amp = mg\sin\theta \\ \amp = 0.5\times 9.81\times \sin\,30^\circ = 2.45\text{ N}. \end{align*}

Therefore,

(b) (i) Spring potential energy with \(\Delta l = 0.12\text{ m}\) gives the answer. Let's refer to this point as b.

(ii) Based on the choice of zero of gravitational potential energy, we have

Therefore, potential energy of the block when it is at this location is

Since the block is at rest here, \(K_\text{b} = 0\text{.}\) Therefore, total energy of the block is

At this point \(U_\text{sp,c} = 0\text{,}\) but we have gravitaional potential energy nonzero.

\begin{equation*} U_\text{gr,b} = mgh_c, \end{equation*}

where \(h_c\) is the vertical height of this point from the point where we choose the zero of the gravitational potential energy. The two points are separated by \(0.12\text{ m}\) along the incline. That means

\begin{equation*} h_c = 0.12\,\sin\,30^\circ = 0.06\text{ m}. \end{equation*}

Therefore,

Using conservation of energy between (b) and this point we get

This gives

Let \(x \) denote the distance on the incline the block moved from the point it was released. This would mean the spring would be extended by

and height \(h_d\) of this point is

Now, setting up conservation of energy with total energy \(0.883\text{ J}\) gives

Solving this, keeping the positive root gives

Strategy: Let \(v_0 \) be the speed of launch and \(v_c\) the speed when the satellite is in the circular orbit. We will first obtain \(v_c\) from Newton's equation of motion for the circular motion. Then, we will get energy of the satellite when in the orbit. After that, energy conservation between energy when in circular orbit and immediately after lauch will give us a condition from which we will obtain \(v_0\text{.}\)

The equation of motion of the circular orbit for uniform circular motion gives

Solve this for \(\dfrac{1}{2}mv_c^2 \) to get

To this we add the potential energy to get the energy \(E_c\) when the satellite is in the circular orbit.

The energy \(E_0\) at the instant after launch is

Equating \(E_0\) and \(E_c\) we get

Therefore,

(a) Since only gravitational force is acting, energy of the spacecraft after launch will be conserved. At the maximum distance point, speed of the craft will be zero.

Thus, we get

Also, note that \(g\) of Earth also relates to its mass and radius by

Using this in Eq. (8.7.5) we get

This can be solved for \(r_\text{max}\text{.}\)

(b) To escape means \(r_\text{max} = \infty\text{.}\) From Eq. (8.7.6), we see that it will occur when

Lets denote this initial speed by \(v_\text{es}\text{.}\)

With \(g=9.81\text{ m/s}^2\text{,}\) \(R_E=6.37\times10^6\text{ m}\text{,}\) we get

(c) The energy needed will be the kinetic energy at the launch site.

Now, we use numerical values to get