\(\displaystyle W=KE_f-KE_i\)

\(\displaystyle W=.5mv_f^2-.5mv_i^2\)

\(\displaystyle W=F_{friction}*d\)

\(\displaystyle F_{friction}=\mu*m*g\)

\(\displaystyle W=\mu*m*g*d\)

Since the car is stopped at the end, \(\displaystyle v_f=0\)

Combining equations:

\(\displaystyle .5mv_i^2=\mu*m*g*d\)

\(\displaystyle .5v_i^2=\mu*g*d\)

Solving for \(\displaystyle d\)

\(\displaystyle \frac{.5v_i^2}{\mu*g}=d\)

From this, it can be seen that doubling the coefficient of friction would cut the stopping distance in half.

\(\displaystyle W=E_f-E_i\)

\(\displaystyle W=.5*50*3^2-.5*50*5^2\)

\(\displaystyle W=-400J\)

\(\displaystyle W=F*d\)

Work is due to friction:

\(\displaystyle W=\mu*m*g*d\)

Solving for \(\displaystyle \mu\)

\(\displaystyle \mu=\frac{W}{mgd}\)

Plugging in values:

\(\displaystyle \mu=\frac{400}{50*9.8*2}\)

\(\displaystyle \mu=.41\)

\(\displaystyle W=E_f-E_i\)

\(\displaystyle W=.5*150*1^2-.5*150*6^2\)

\(\displaystyle W=-2663J\)

\(\displaystyle W=F*d\)

Work is due to friction:

\(\displaystyle W=\mu*m*g*d\)

Solving for \(\displaystyle \mu\)

\(\displaystyle \mu=\frac{W}{mgd}\)

Plugging in values:

\(\displaystyle \mu=\frac{-2663}{150*-9.8*3.5}\)

\(\displaystyle \mu=.51\)

Because the block is traveling on an inclined ramp and experiences friction, the two main forces at work here are a component of gravitational force and kinetic frictional force.

 The total force acting on the block can be written as \(\displaystyle F_{total} = F_{g, \parallel } + f_{k}\), where \(\displaystyle F_{g, \parallel }\) is the component of gravitational force in the direction of motion, and \(\displaystyle f_{k}\) is the kinetic frictional force.

We know that forces can be expressed as \(\displaystyle F=ma\), and kinetic friction can be expressed as \(\displaystyle f_{k} = -\mu_k N\). Hence our total force can be expressed as \(\displaystyle F_{total} = ma_{total} = ma_{g, \parallel } - \mu_{k}N\). The normal force can be further expressed as \(\displaystyle N = ma_{g, \perp }\)

The component of gravity parallel to the ramp is given by \(\displaystyle a_{g,\parallel} = a_{total}sin(20\degree)\), while the component of gravity perpendicular to the ramp is given by \(\displaystyle a_{g, \perp }=a_{total}cos(20\degree)\).

Our total force can hence be expressed as \(\displaystyle ma_{total} = ma_{g, \parallel } - \mu_{k}ma_{g, \perp }\).

Thus, \(\displaystyle \mu_k = \frac{ma_{g, \parallel } - ma_{total}}{ma_{g, \perp }} = \frac{a_{total}sin(20 \degree) - a_{total}}{a_{total}cos(20\degree)}\)

and \(\displaystyle \mu_k = \frac{(-5\frac{m}{s^2})sin(20\degree) - (-5 \frac{m}{s^2})}{(-5\frac{m}{s^2})cos(20\degree)} = 1.43\)

\(\displaystyle F_{net}=ma=F_1+F_2\)

\(\displaystyle ma=F_{rocket}+F_{friction}\)

\(\displaystyle ma=F_{rocket}-\mu_kmg\)

Solving for the coefficient of friction:

\(\displaystyle \frac{F_{rocket}-ma}{mg}=\mu_k\)

\(\displaystyle \frac{3500-(170+70)*10}{(70+170)*9.8}=\mu_k\)

\(\displaystyle \mu_k=.47\)

We can use the expression for conservation of energy to solve this problem:

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

The problem statement tells us that the sledder is initially at rest, so we can eliminate initial kinetic energy. Also, if we assume that the bottom of the hill has a height of 0, we can eliminate final potential energy to get:

\(\displaystyle U_i + W= K_f\)  (1)

From here we need to determine what is going to contribute to work. The only extra piece of information we have in this problem is that there is an average frictional force. Therefore that will be the only source of work. Let's began expanding each term, going from left to right:

\(\displaystyle U_i = mgh\)  (2)

Note that we didn't mark the height as an initial height because we assumed that the bottom of the hill has a height of 0.

\(\displaystyle W= F_{f_{avg}}d\)

Where d is the length of the hill. We can calculate this distance using the height and angle of the hill.

\(\displaystyle sin(a)= \frac{h}{d}\)

Rearranging for distance:

\(\displaystyle d = \frac{h}{sin(a)}\)

Substituting this back into our expression for work, we get:

\(\displaystyle W = F_{f_{avg}}\left (\frac{h}{sin(a)} \right )\)  (3)

Now our final term:

\(\displaystyle K_f = \frac{1}{2}mv_f^2\)    (4)

Now substituting expression 2, 3, and 4 into expression 1, we get:

\(\displaystyle mgh - F_{f_{avg}}\left (\frac{h}{sin(a)} \right ) = \frac{1}{2}mv_f^2\)

The reason we are subtracting friction is because it is removing energy from the system. Another way we could have written our initial expression would be to have work on the final state and then friction would be positive.

Rearranging for the frictional force:

\(\displaystyle F_{f_{avf}}\left (\frac{h}{sin(a)} \right )= \frac{2mgh-mv_f^2}{2}\)

\(\displaystyle F_{f_{avg}} = \frac{\left ( 2mgh-mv_f^2\right )sin(a)}{2h}\)

We have all of these values, so time to plug and chug:

\(\displaystyle F_{f_{avg}} = \frac{\left ( 2(75kg)\left ( 10\frac{m}{s^2}\right )(10m)-(75kg)\left ( 9\frac{m}{s}\right )^2\right )sin(30^{\circ})}{2(10m)}\)

\(\displaystyle F_{f_{avg}} = 223 N\)

You simply need to know the formula for the potential energy stored in a spring to solve this problem. The formula is:

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

where k is the spring constant, and x is the distance from equilibrium

We can rearrange this to get:

\(\displaystyle x = \sqrt{\frac{2U}{k}}\)

Plugging in our values, we get:

\(\displaystyle x = \sqrt{\frac{2\cdot20J}{15\frac{N}{m}}} = \sqrt{2.\overline{66}m^2} = 1.63m\)

Since the block is motionless, we know that our forces will cancel out:

\(\displaystyle F_{net}=0\)

There are three forces in play: one from each spring, as well as the force of gravity. If we assume that forces pointing up are positive, we can write:

\(\displaystyle F_{spring,top}+F_{spring,bot} - mg = 0\)

Plugging in expressions for each spring force, we get:

\(\displaystyle kx_{top}+kx_{bot} - mg = 0\)

Rearring for the displacement of the top spring, we get:

\(\displaystyle x_{top}=\frac{mg-kx_{bot}}{k} = \frac{(3kg)(10\frac{m}{s^2})-(25\frac{N}{m})(0.4m)}{25\frac{N}{m}}\)

\(\displaystyle x_{top} = \frac{30N - 10N}{25\frac{N}{m}}= 0.8m\)

Since the block is motionless, we can assume that the forces cancel out:

\(\displaystyle F_{net}=0\)

If we designate any forces pointing downward as positive, we can write:

\(\displaystyle F_g +F_{spring,bot}-F_{spring,top}=0\)

Inserting expressions for each force, we get:

\(\displaystyle mg + kx_{bot}- kx_{top} = 0\)

Rearranging for mass, we get:

\(\displaystyle m = \frac{k(x_{top}-x_{bot})}{g} = \frac{10\frac{N}{m}(2.5m-0.75m)}{10\frac{m}{s^2}}\)

\(\displaystyle m = 1.75kg\)

There are two ways to solve this problem: the first involves calculating the spring constant and the second does not. We'll go through both methods.

Calculating Spring Constant

We can use the expression for the force of a spring:

\(\displaystyle F=kx\)

At equilibrium, the force of the spring equals the force of gravity:

\(\displaystyle mg = kx\)

Rearranging for the spring constant and plugging in values, we get:

\(\displaystyle k = \frac{mg}{x}=\frac{(2kg)(10\frac{m}{s^2})}{0.2m}= 100\frac{N}{m}\)

Now, apply this equation when the spring is on a different planet:

\(\displaystyle mg = kx\)

Rearranging for displacement and plugging in values, we get:

\(\displaystyle x = \frac{mg}{k}=\frac{(2kg)(4\frac{m}{s^2})}{100\frac{N}{m}} = 0.08m\)

Without Calculating Spring Constant

We can write the force equation for each scenario. Let the subscript 1 denote Earth, and 2 denote the other planet:

\(\displaystyle mg_1=kx_1\)

\(\displaystyle mg_2 = kx_2\)

Using these equations, we can set up a proportion:

\(\displaystyle \frac{mg_2}{mg_1} = \frac{kx_2}{kx_1}\)

\(\displaystyle \frac{g_2}{g_1}=\frac{x_2}{x_1}\)

Rearranging for the displacement of scenario 2 and plugging in values:

\(\displaystyle x_2 = x_1\left ( \frac{g_2}{g_1}\right ) = (0.2m)\left ( \frac{4\frac{m}{s^2}}{10\frac{m}{s^2}}\right )=0.08m\)