Using conservation of energy:

\(\displaystyle KE_1+EPE_1=KE_2+EPE_2\)

\(\displaystyle .5mv_1^2+.5kx_1^2=.5mv_2^2+.5kx^2\)

Plugging in values:

\(\displaystyle .5.050*2^2+.5k(0)^2=.5*.050(0)^2+.5k*.05^2\)

Solving for \(\displaystyle k\)

\(\displaystyle k=\frac{80N}{m}\)

This question covers the conservation of energy:

\(\displaystyle E = U_{i}+K_{i}=U_{f}+K_{f}\)

From the problem statement, we know that the cart has both initial potential and kinetic energy. We can assume that there is no final potential energy. This means that there is final kinetic energy from the final velocity that we are trying to find.

After eliminating final potential energy, we can rewrite the equation as:

\(\displaystyle mgh_{i} + \frac{1}{2}mv_{i}^2 = \frac{1}{2}mv_{f}^2\)

We can eliminate mass from the equation to get:

\(\displaystyle gh_{i} + \frac{1}{2}v_{i}^2 = \frac{1}{2}v_{f}^2\)

We are solving for final velocity, so rearranging, we get:

\(\displaystyle v_{f}= \sqrt{2gh_{i}+v_i^2}\)

Values for each variable were given in the problem statement, so we can simply plug in these values to solve:

\(\displaystyle v_{f}= \sqrt{2(10)(30)+2^2} = \sqrt {604} = 24.6 \frac{m}{s}\)

According to the problem statement, we can treat the friends as a single mass of 500kg. We can use the expression for conservation of energy to calculate the velocity as they hit the trampoline.

\(\displaystyle E = U_i+K_i = U_f+K_f\)

If we take the initial state to be when the friends are at their highest point and the final state to be when the friends hit the trampoline, we can rewrite:

\(\displaystyle U_i = K_f\)

Substituting in expressions:

\(\displaystyle mgh_i=\frac{1}{2}mv_f^2\)

Rearranging for velocity:

\(\displaystyle v_f = \sqrt{2gh_i}\)

Plugging in our values:

\(\displaystyle v_f = \sqrt{2(10\frac{m}{s^2})(2m)} = 6.32 \frac{m}{s}\)

The statement says that the friends decelerate from this velocity to zero over a period of 0.5 seconds. Therefore, we can write:

\(\displaystyle a = \frac{\Delta v}{t} = \frac{6.32\frac{m}{s}}{0.5s} = 12.65 \frac{m}{s^2}\)

We can use this to calculate the force that the trampoline is exerting on the friends:

\(\displaystyle F = ma = (500kg)\left ( 12.65\frac{m}{s^2} \right ) = 6325 N\)

This is above the threshold of 5000 N, so the trampoline does break.

There are two ways to solve this problem. The first uses the concept of conservation of energy, and the second uses kinematics. We'll go through both methods, as you should be comfortable using either one, as some problems won't have the ability to be easily solved using two different methods.

Method 1: Conservation of Energy

Note: The only reason we can use this method is because we know the height at which the ball crosses over the center field wall. Without that height, we would have to do method 2.

We will first split the initial velocity of the ball into its components:

\(\displaystyle v_{xi} = 40cos(55^{\circ}) = 22.9 \frac{m}{s}\)

\(\displaystyle v_{yi} = 40sin(55^{\circ}) = 32.8\frac{m}{s}\)

Since we are neglecting air resistance, the x-component stays constant. Therefore we can say:

\(\displaystyle v_{xf} = 22.9 \frac{m}{s}\)

We can find the final y-componenet using our conservation of energy equation:

\(\displaystyle E = U_i+K_i=U_f+K_f\)

Substituting in our expressions:

\(\displaystyle mgh_i + \frac{1}{2}mv_i^2 = mgh_f + \frac{1}{2}mv_f^2\)

Canceling out mass and rearranging for final velocity:

\(\displaystyle g(h_i-h_f)+\frac{1}{2}v_i^2 = \frac{1}{2}v_f^2\)

\(\displaystyle v_f = \sqrt{2g(h_i-h_f)+v_i^2}\)

Note that this is one of the big five kineamtics equations with which you should be familiar. Plugging in our values, we get:

 \(\displaystyle v_{f}= \sqrt{2(10\frac{m}{s^2})(1m - 33m)+(32.8\frac{m}{s})^2} = \sqrt{-640\frac{m^2}{s^2}+1073.616\frac{m^2}{s^2}}\)

\(\displaystyle v_{fy} = 20.8 \frac{m}{s}\)

Now that we have the components, we can combine them to get the final velocity:

\(\displaystyle v_f = \sqrt{v_{fx}^2 + v_{fy}^2} = \sqrt{(22.9)^2 + (20.8)^2}\)

\(\displaystyle v_f = 31.0 \frac{m}{s}\)

Method 2: Kinematics

We start off the same way by separating the velocities into their components:

\(\displaystyle v_{xi} = 40cos(55^{\circ}) = 22.9 \frac{m}{s}\)

\(\displaystyle v_{yi} = 40sin(55^{\circ}) = 32.8\frac{m}{s}\)

Since we are neglecting air resistance, the x-component stays constant. Therefore we can say:

\(\displaystyle v_{xf} = 22.9 \frac{m}{s}\)

We also know how far the ball travels horizontally, so we can find out long it takes to cover that distance:

\(\displaystyle t = \frac{d}{v_x} = \frac{123 m}{22.9 \frac{m}{s}} = 5.36 s\)

We can then use this time to find out the final vertical velocity:

\(\displaystyle v_{fy} = v_{fi} - at\)

The acceleration is subtracted because the acceleration is technically negative.

\(\displaystyle v_{fy} = 32.8 \frac{m}{s}- (10\frac{m}{s^2})(5.36s) =-20.8 \frac{m}{s}\)

The velocity is negative because the ball is now traveling back downward. We can now combine the component velocities into the final velocity of the ball:

\(\displaystyle v_f = \sqrt{v_{fx}^2 + v_{fy}^2} = \sqrt{(22.9)^2 + (20.8)^2}\)

\(\displaystyle v_f = 31.0 \frac{m}{s}\)

Each time the ball bounces, 10% of the kinetic energy is lost. This means that after colliding with the ground, the ball has 90% of its previous potential energy at the top of its flight. To find the total energy after four bounces, we multiply the initial potential energy by \(\displaystyle \small 0.9^4\).

The initial potential energy is:

\(\displaystyle PE=mgh\)

\(\displaystyle PE=(0.25kg)(9.8\frac{m}{s^2})(10m)\)

\(\displaystyle PE=24.5J\)

Use our exponential expression to find the remaining energy after four bounces.

\(\displaystyle KE_4=24.5*(0.9)^4\)

\(\displaystyle (24.5J)*(0.66)=16.1J\)

Based on conservation of energy, we can find the height that the ball travels with this remaining kinetic energy

\(\displaystyle PE_4=16.1J=mgh_4\)

\(\displaystyle h_4=\frac{PE_4}{mg}\)

\(\displaystyle h_4=\frac{16.1J}{(0.25kg*9.8\frac{m}{s^2}}=6.6m\)

The energies involved in this problem are kinetic and potential energy. Conservation of energy shows that the initial energies will be equal to the final energies.

\(\displaystyle mgh_i+\frac{1}{2}mv_i^2=mgh_f+\frac{1}{2}mv_f^2\)

Choosing the bottom of the incline to be the zero height, the ball starts out with kinetic energy and zero potential energy. When the ball reaches maximum height, its velocity is zero (zero kinetic energy). This simplifies our energy equation.

\(\displaystyle 0+\frac{1}{2}mv_i^2=mgh_f+0\)

Isolate the height variable and use the given values to solve for the maximum height.

\(\displaystyle h_f=\frac{v_i^2}{2g}\)

\(\displaystyle h_f=\frac{(17\frac{m}{s})^2}{2(9.8\frac{m}{s^2})}\)

\(\displaystyle h_f=14.7\ \text{m}\)

This is the vertical height. The work done by gravity is calculated as the product for force and distance.

\(\displaystyle W=F_{gravity}h_f=-mgh_f\)

The minus sign indicates that the force of gravity acts downward (negative direction).

\(\displaystyle W=-(2.7kg)(9.8\frac{m}{s^2})(14.7m)\)

\(\displaystyle W=-389\ \text{J}\)

This problem covers the conservation of energy (including friction):

\(\displaystyle E = U_i +K_i = U_f+K_f + F\)

The only term we can eliminate is final potential energy:

\(\displaystyle U_i +K_i - F = K_f\)

Expanding each of the terms, we get:

\(\displaystyle mgh +\frac{1}{2}mv_i^2 - F = \frac{1}{2}mv_f^2\)

We can rearrange this to solve for final velocity:

\(\displaystyle v_f = \sqrt{2gh +v_i^2 - \frac{2F}{m}}\)

Many students will look at this and feel that the equation has just become more complex or harder to follow. However, the idea behind this is to reduce calculation errors. By doing all of your subsitutions of variables and rearrangements of the equations before plugging in your values, you only have to do a single calculation. This greatly reduces the frequency of silly calculation errors. Furthermore, it tends to make it much easier to follow your units in case you need to troubleshoot the problem.

Now, plugging in all of our values, we get:

\(\displaystyle v_f = \sqrt{2(10\frac{m}{s^2})(15m)+(4\frac{m}{s})^2-\frac{2(5000J)}{500kg}}\)

\(\displaystyle v_f = \sqrt{300\frac{m^2}{s^2}+16\frac{m^2}{s^2} - 20 \frac{m^2}{s^2}}\)

\(\displaystyle v_f = 17.2 \frac{m}{s}\)

First, we'll write out the equation for conservation of energy:

\(\displaystyle E + U_i+K_i=U_f+K_f+W_f\)

The velocity of the block is constant, so kinetic energies will cancel out. Furthermore, we can remove final potential energy if we say that the final height is 0m. Therefore, we get:

\(\displaystyle U_i = W_f\)

Substituting in the expression for potential energy, we get:

\(\displaystyle W_f = mgh_i = (10kg)(10\frac{m}{s^2})(10m) = 1kJ\)

Note that this problem cannot be solved by using the force of gravity and force of friction because we do not know the angle of the slope.

Energy in a system must be conserved, and the energy in the falling pebble is either potential or kinetic:

\(\displaystyle mgh_1+\frac{1}{2}mv_1^2=mgh_2+\frac{1}{2}mv_2^2\)

Potential energy is represented by the \(\displaystyle mgh\) terms.

Kinetic energy is represented by the \(\displaystyle \frac{1}{2}mv^2\) terms.

At the exact moment the pebble is released, its stationary, so \(\displaystyle v_1=0\)

\(\displaystyle mgh_1=mgh_2+\frac{1}{2}mv_2^2\)

\(\displaystyle \frac{1}{2}mv_2^2=mg(h_1-h_2)\)

\(\displaystyle v_2=\sqrt{2g(h_1-h_2)}\)

\(\displaystyle v_2=\sqrt{2(9.81\frac{m}{s^2})(22m-2m)}\)

\(\displaystyle v_2=19.81\frac{m}{s}\)

This problem deals with both potential energy and kinetic energy.

Potential energy is expressed as:

\(\displaystyle PE=mgh\)

Kinetic energy is expressed as:

\(\displaystyle KE=\frac{1}{2}mv^2\)

Energy must be conserved, so set up the following equation:

\(\displaystyle mgh_1+0.5mv_1^2=mgh_2+0.5mv_2^2\)

The initial height can be treated as zero, as can the final velocity. Plug in these zero values into the above equation.

\(\displaystyle 0.5mv_1^2=mgh_2\)

Solve for \(\displaystyle v_1\), the initial velocity the ball needs in order to reach a maximum height of \(\displaystyle 50m\).

\(\displaystyle v_1^2=2gh_2\)

\(\displaystyle v_1=\sqrt{2gh_2}\)

\(\displaystyle v_1=\sqrt{2\left(9.81\frac{m}{s^2}\right)(50m)}\)

\(\displaystyle v_1=31.32\frac{m}{s}\)