Detailed Answer

This is a typical energy conservation question, so we need to take the energy at one point, and have it equal the energy at another point. Its logical to take one of these points when the spring is fully compressed, and the block isnt moving yet, and the other point when the block is just at the top of the slope. In the first point we only have elastic energy, which is given by \(\dfrac{1}{2} kx^2\), and at the top we have potential, and kinetic energy. Setting these 2 equal gives: \(\dfrac{1}{2} kx^2 = mgh + \dfrac{1}{2} mv^2\). Plugging in, \(\dfrac{1}{2} k(0.1)^2 = (2)(10)(6) + \dfrac{1}{2} (2)(10)^2\), yeilds B. Note: we can use g = 10, to make calculations easier. If the angle were given we could also calculate the initial height it is starting from, so we could also calculate its initial potential energy.

Detailed Answer

First we should calculate the acceleration of the car over this period which we can do with the equation \(50^2 = 0^2 + 2a(200)\) which gives \(6.25 ms^{-2}\). Then we can calculate the time it was accelerating for by doing \(200 = \dfrac{1}{2}(6.25)(t^2)\) gives that t = 8s. Then we need to calculate the kinetic energy that the car has gained through this acceleration. Which is: \(\dfrac{1}{2}(1000)(50)^2 = (1.25)(10^6) J\). Since we know \(Power = \dfrac{Energy}{Time}\) we can do \(P = \dfrac{1.25(10^6)}{8} = 156250 W\) .

Detailed Answer

Efficiency is the ratio of useful power to the input power. So \(0.6 = \dfrac{Useful Power}{50}\) which gives 30W for the useful power.

Detailed Answer

Work is force times distance. If we multiply the y axis with the mass M, then we get the force developed by the car as a function of distance. We know the area under a force distance graph is the work done by the engine, hence C.

Detailed Answer

The energy stored in a spring is given by \(E = \dfrac{1}{2}kx^2\). Plugging in the values for the new spring \(E_2 = \dfrac{1}{2}(5k)(\dfrac{1}{2}x)^2 = \dfrac{5}{4} * \dfrac{1}{2}kx^2 = \dfrac{5}{4}E \), hence B.

Answer and Explanation

To convert from kWh to joules we need to multiply by \( 3.6 \cdot 10^6 \), as 1 hour is 3600 seconds and the kiloWatt is 1000 watts.

Answer and Explanation

In an inelastic collision kinetic energy is not conserved. Momentum is always conserved as long as there are no external forces acting, which here there aren't.

Explanation

We need to use the formula \( E = \frac{p^2}{2m} \). Since \( p_1 = p_2 \), if we rearrange the first equation for \( p \), we can set them equal to each other.

\[ p^2_1 = p^2_2 = 2m_1E_1 = 2m_2E_2 \]So: \[ E_2 = \frac{m_1}{m_2} E_1 = \frac{1}{2}E_1 \]

Explanation

Power is energy over time. The frog gained a potential energy of \( mgh \) in a time \( \Delta t \), so the power is:

\[ P = \frac{E}{t} = \frac{mgh}{\Delta t} \]

Explanation

Since no external forces act, momentum must be conserved. However, kinetic energy cannot be conserved as there is a frictional force acting inside the watermelon, slowing the bullet down and converting its kinetic energy into thermal energy.

Explanation

The gravitational force always acts perpendicular to Earth's velocity, which is tangential to the circular orbit. The work done is given by \( W = F s \cos{\theta} \), where \( \theta \) is the angle between the force and displacement. Since the angle is always 90°, \( \cos{90^\circ} = 0 \), the work done is zero.

Explanation

When we connect the springs in series, the new spring constant will be \( \frac{k}{2} \). Since they are connected in series, they both experience the same force, and therefore, both springs will have a displacement \( x \). This results in a total displacement of \( 2x \) because both springs are compressed. The new energy can be calculated as:

\[ E_{series} = \frac{1}{2} k_{\text{new}} x_{\text{new}}^2 = \frac{1}{2} \cdot \frac{k}{2} (2x)^2 = k x^2 \] This is double the initial energy of \(\frac{1}{2} k x^2\).

Explanation

Mechanical energy is conserved, so when kinetic energy is a maximum, potential energy is at a minimum after which the potential energy will start to increase at the expense of the kinetic energy. Thus at some point, they must be equal. They will be equal once between the 0 displacement line and the maximum displacement line, and also once between the 0 displacement line and the minimum displacement line. Since we considered a full oscillation, this happens twice, so the answer is 4.