Conservative forces are forces that do not lose energy to heat, sound, or light. Of these answers, energy is completely conserved and transferred from kinetic energy to potential energy, or vice versa. Gravitational forces, electrostatic forces, and elastic forces all work by providing a potential that will work in the same direction as the motion of an object or particle, allowing kinetic and potential energy to interconvert. Frictional forces lose energy as heat when sliding across a surface, and the more force (the more rough the surface), the more energy that is lost.

One-half of the force will be required to pull the platform up in the new scenario. In the diagram below, now notice that the box has two upward tension forces due to the pulley attached directly to the box. These two tension forces act against the weight of the box, meaning that each tension force is one half of the weight of the box. Because tension is transmitted unchanged over the length of the rope, the tension a person would need to pull against would be half the weight of the box. This leads us to determine that half the force would be required to lift the box if two pulleys were used.

The question states that:

\(\displaystyle F_z=\frac{1}{3}F_x\)

Recall that the passage states that \(\displaystyle F_y\) is also one third of \(\displaystyle F_x\).

\(\displaystyle F_y=\frac{1}{3}F_x\)

The force exerted by student Y and student Z is the same.

\(\displaystyle F_y=F_z\)

Mechanical advantage is defined as:

\(\displaystyle \text{Mechanical advantage}=\frac{\text{Object weight}}{\text{Force to lift object}}\)

The weight of the object is the same for both students because both of them are lifting the same boulder. The force applied is also the same; therefore, mechanical advantage for both machines is the same.

The definition of mechanical advantage is:

\(\displaystyle \text{Mechanical advantage}=\frac{\text{Object weight}}{\text{Force to lift object}}\)

Rearranging this equation and solving for weight gives:

\(\displaystyle \text{Object weight}=(\text{Mechanical advantage})(\text{Force to lift object})\)

\(\displaystyle \text{Object weight}=(5)(6N)=30N\)

Mechanical advantage is a unitless quantity. Remember the difference between weight and mass. Weight is a measure of force and has units of Newtons, whereas mass has units of kilograms.

Conservation of energy is the key here. Initial energy is all gravitational potential energy:

\(\displaystyle E_{i} = mg(h + x)\) 

Note that the final height change is equal to the height above the spring added to the displacement of the spring.

This is equal to the final energy, which is all spring potential energy:

\(\displaystyle E_{f}=\frac{1}{2}kx^{2}\)

Set these equations equal and solve for the spring constant.

\(\displaystyle \frac{1}{2}kx^{2}= mg(h + x)\)

\(\displaystyle k=\frac{2mg(h+x)}{x^2}\)

\(\displaystyle k=\frac{2(2kg)(9.8\frac{m}{s})(3m+0.20m)}{(0.20m)^2}\)

\(\displaystyle k=3136\frac{N}{m}\)

The density of an object is equal to mass over volume.

\(\displaystyle \rho=\frac{m}{V}\)

Knowing the length and the cross sectional area of the log, we can find its volume.

\(\displaystyle V=(area)(length)=(5cm^2)(25cm)=125cm^3\)

Plugging in volume and mass into the equation will enable us to find density.

\(\displaystyle \rho=\frac{100g}{125cm^3}=0.80\frac{g}{cm^3}\)

The formula for density is:

\(\displaystyle \rho=\frac{mass}{volume}\)

In the question, we are given the densities of both solids and a means to find their volumes. Using these values, we will be able to determine the mass of each solid.

\(\displaystyle m=\rho*V\)

\(\displaystyle m_A=(0.25\frac{g}{cm^3})(36cm^3)=9g\)

\(\displaystyle m_B=(0.10\frac{g}{cm^3})(5cm)^3=12.5g\)

Now that we know both masses, we can find the difference.

\(\displaystyle m_B-m_A=12.5g=9g=3.5g\)

Solution A and solution X have the same concentration, therefore, we are only concerned with temperature differences between the two solutions. Density is dependent on temperature: as temperature increases density decreases. Recall the definition of density:

\(\displaystyle \rho=\frac{\text{mass}}{\text{volume}}\)

Increasing the temperature will slightly increase the volume of the solution and, subsequently, decrease density. The temperature has no effect on mass. The solution at the higher temperature (solution X) will have a lower density.

Plotted in this way, the slope of the line is equal to the density of the substance.

\(\displaystyle \rho=\frac{\text{mass}}{\text{volume}}=\frac{g}{mL}\)

We would expect the line to be linear and positive. A slope close to zero could indicate that the data was not graphed optimally or that the density is extremely small, and a density greater than one is possible (anything that sinks in water has a density greater than one). A negative density would indicate that greater masses of this substance have smaller volumes, which, for a pure substance, does not make sense.

Solving this question requires that we use the given densities to convert mass to volume.

\(\displaystyle 6.0 lb H_{2}O * \frac{454 g}{lb} * \frac{1 mL}{1.0 g}* \frac{1 L}{1000 mL} = 2.7 L\)

\(\displaystyle 50 lb Pb * \frac{454 g}{lb} * \frac{1 mL}{11.4 g}* \frac{1 L}{1000 mL} = 2.0 L\)

\(\displaystyle 19 lb Al * \frac{454 g}{lb} * \frac{1 mL}{2.7 g}* \frac{1 L}{1000 mL} = 3.2 L\)

We see that only the aluminum sample has a volume over three liters.