Known

\(\displaystyle m = 1.90kg\)

\(\displaystyle W = 75J\)

Unknown

\(\displaystyle h = ?\)

When something does work on an object it changes the total energy of the object.  In this case, the work done by the person converts to kinetic energy as the stone is launched.  This kinetic energy is then turned into gravitational potential energy when the stone is at the highest point of the peak.

Work done = Kinetic Energy when the stone leaves the hand = Gravitational Potential Energy at the peak

\(\displaystyle GPE = mgh\)

We can set the work done to the gravitational potential energy

\(\displaystyle W = mgh\)

Plug in our known values and solve.

\(\displaystyle 75J = (1.90kg)(9.8m/s^2)h\)

\(\displaystyle h = 4.03m\)

Known

\(\displaystyle v = 60km/h\)

\(\displaystyle m = 1100kg\)

\(\displaystyle \Delta x = 2.0m\)

We need to convert our velocity to \(\displaystyle m/s\)

\(\displaystyle 60km/h * 1000m/1km * 1h/60min *1 min/60sec = 16.67m/s\)

Unkonwn

\(\displaystyle k = ?\)

The easiest way to solve this problem is through conservation of energy.  When the car is rolling along the road it has kinetic energy.  Once the spring brings it to a complete stop, the car has elastic potential energy. According to the law of conservation of energy, these two values should be equal to one another

\(\displaystyle Kinetic \, Energy \, Initial = Spring \, Potential \, Energy \, Final\)

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

\(\displaystyle PE_{spring} = \frac{1}{2} k\Delta x^2\)

\(\displaystyle \frac{1}{2} mv^2 = \frac{1}{2}k\Delta x^2\)

We can now plug in our known values and solve for the missing variable

\(\displaystyle \frac{1}{2} (1100kg)(16.67m/s)^2 = \frac{1}{2} (k)(2m)^2\)

\(\displaystyle 15838.895J = 2k\)

\(\displaystyle K = 7,919 N/m\)

The equation for potential energy is \(\displaystyle PE=mgh\). We are given the mass of the ball, the height of the table, and the acceleration of gravity in the question. The distance the ball travels is in the downward direction, making it negative.

Plug in the values, and solve for the potential energy.

\(\displaystyle PE = (0.5kg)(-9.8m/s^2)(-1.5m)\)

\(\displaystyle PE=7.35J\)

The units for energy are Joules.

Knowns

\(\displaystyle PE = 25J\)

\(\displaystyle k =500N/m\)

Unknowns

\(\displaystyle x = ?\)

Equation

\(\displaystyle PE =\frac{1}{2} k x^2\)

Plug in the vales and solve for the stretch of the spring.

\(\displaystyle 25J = \frac{1}{2} (500N/m)x^2\)

\(\displaystyle x^2= 0.1\)

\(\displaystyle x = 0.32m\)

Work causes a change in the kinetic energy of an object.  In the example of a car stopping, the work done on the car causes the car to slow down to a stop, therefore changing the kinetic energy.

Work is also equal to the force times the displacement of the object. In this case, we are assuming that the force applied to stop the car does not change.

\(\displaystyle W = Fd\)

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

\(\displaystyle -W = \Delta KE\)

\(\displaystyle -W = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2\)

Since the car is coming to a stop the final velocity of the car is 0m/s.

\(\displaystyle -W = - \frac{1}{2} mv_i^2\)

\(\displaystyle W = \frac{1}{2} mv_i^2\)

\(\displaystyle Fd =\frac{1}{2} mv_i^2\)

Therefore when you double the velocity, that value is then squared.

\(\displaystyle 2^2 = 4\)

Therefore your kinetic energy is increased by \(\displaystyle 4\) times the original amount.

If your kinetic energy is \(\displaystyle 4\) times greater, than with the same force being applied, the stopping distance will also increase by \(\displaystyle 4\) times since they are directly related.

Hooke’s law states that spring constant is directly related to the force applied and the distance that the object is stretched.

\(\displaystyle k = F/\Delta x\)

We also know that the potential energy of a spring is related to the spring constant and the distance that the object is stretched

\(\displaystyle PE_{spring} = \frac{1}{2} k \Delta x^2\)

We can substitute our equation for Hooke’s law into the potential energy equation.

\(\displaystyle PE_{spring} = \frac{1}{2} (F/\Delta x) \Delta x^2\)

This simplifies to

\(\displaystyle PE_{spring} = \frac{1}{2} F\Delta x\)

This equation shows that there is a direct relationship between the force on the spring and the potential energy of the spring.  If the force is doubled, then the potential energy will likewise double.

Known

\(\displaystyle k = 55N/m\)

\(\displaystyle m = 3kg\)

\(\displaystyle x_i = 12cm = 0.12m\)

Unknown

\(\displaystyle \Delta x\)

First let us consider the force acting on the spring. Since the spring is hanging vertically, the only force acting on the spring is the force of gravity from the mass that has been added to the spring.

\(\displaystyle F_G = mg\)

We can substitute in our variables and find the gravitational force.

\(\displaystyle F_G = (3kg)(9.8m/s^2)\)

\(\displaystyle F_G = 29.4 N\)

According to Hooke’s Law the spring constant of a spring is directly proportional to the force acting on it and inversely proportional to the amount of stretch.

\(\displaystyle k = F/\Delta x\)

We can use the provided spring constant and the force acting on the spring to determine the amount of stretch.

\(\displaystyle 55N/m = 29.4N/\Delta x\)

\(\displaystyle \Delta x = 0.53m\)

\(\displaystyle \Delta x = x_f -x_i\)

\(\displaystyle 0.53m = x_f - 0.12m\)

\(\displaystyle x_f = 0.65m\)

To solve for the kinetic energy, we will need to use the equation:

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

Before we can plug in our given values, we must convert the mass from grams to kilograms. Remember, the SI unit for mass is kilograms, so most calculations will require this conversion.

\(\displaystyle 145g * \frac{1kg}{1000g} = 0.145kg\)

Now we can use this mass and the given velocity to solve for the kinetic energy.

\(\displaystyle KE = \frac{1}{2}(0.145kg)(38m/s)^2\)

\(\displaystyle KE = \frac{1}{2}(0.145kg)(1444m^2/s^2)\)

\(\displaystyle KE = 104.69J\)

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

Therefore when you double the velocity, that value is then squared.

\(\displaystyle 2^2 = 4\)

Therefore your kinetic energy is increased by 4 times the original amount.