Recall the formula to determine kinetic friction:

Here,  is the kinetic friction force,  is the constant of friction, and  is the magnitude of the normal force of the object in question. Recall that the formula for the normal force is given by:

Here,  is the mass of the object,  is the gravitational constant, and  is the angle of elevation. Since the object is lying  to the horizontal, the normal force it feels is:

Therefore,

To solve for the units, we simply have to solve for  and determine its units. 

Both friction and normal forces have units of newtons , which means that  has units . 

Therefore,  is a dimensionless term and has no units. 

The equation for frictional force is given by:

Because the crate is on a horizontal platform, the normal force is equal to the weight force . Multiply the normal force by the given coefficient of kinetic friction  and solve for the frictional force:

Note that the coefficient of friction is unitless because it is a ratio and therefore our answer comes out in Newtons. Also note that the force of friction is in a direction that opposes motion.

We can use one of the kinematic equations to our advantage: 

We are given the player's speed and the distance the player slide into home plate. What is implied is that the player's final velocity . We can also rewrite the acceleration  in terms of the player's mass and the forces acting on him (which is only friction)

Substitute this into the kinematic equation and solve for the coefficient of kinetic friction, 

Since the car is stopped at the end:

Solve for coefficient of friction:

Plug in values:

Use the following equations:

The force of friction will point in the opposite direction as the direction of travel, so it will have a negative sign.

Since the energy will be zero when the vehicle has stopped.

Solve for , the stopping distance.

, the mass is not in the solution. Indeed, the extra energy due to mass is cancelled out due to extra friction resulting from the higher mass.

The diagram above shows the free body diagram for the suitcase. For this problem, we take the positive y-direction to be the direction of the normal force and the negative x-direction to be the direction of the frictional force. Then applying Newton's 2nd law in the x and y-directions gives us the following equations:

Since the child pulls the suitcase at constant velocity, we have  by Newton's first law. Also, the object is constrained to the ground, which means that  as well. Thus we obtain:

We want to solve for , but we need to get rid of the normal force in the first equation. We can do so by solving for  in the second equation. Doing so, we obtain:

Substituting this value into the first equation and solving for , we obtain:

Substituting our unknowns into the expression on the right side, we get:

Therefore, the mass of the suitcase is 

In this question, we're presented with a scenario in which we need to consider the frictional forces that occur between the tire of a moving vehicle and the road. More specifically, we'll need to determine the direction in which the frictional force acts, as well as whether the force is kinetic or static.

First, let's determine what type of force we're dealing with. Since the tire is rotating as the car itself moves forward, each point around the circumference of the tire will come into contact with the road. It's important to realize that since the surface of the tire is never moving relative to the road in which it is in contact with, this cannot be kinetic friction. Therefore, it is the static force of friction that the road provides in moving the car forward.

Next, let's see which direction the force is acting. If we imagine the car moving to the right towards A, we see that the car's tire is pushing against the road to the left. And because of Newton's third law, since the tire is pushing against the road towards the left, the road must be pushing back with an equal magnitude of force, but in the opposite direction, to the right. Thus, the frictional force of the road is both static and acting toward the right.

We will begin with Newton's second law to solve this problem:

There are two forces acting on the sledder in the direction of acceleration: gravity and friction. We will begin with gravity:

However, this is a vertical force, and we need the component that is in the direction of acceleration. Therefore, we will multiply it by the sine function:

If you're wondering why sin was used, think about the situation practically. As the angle grows, the hill gets steeper, and there will be more gravitational force in the direction of acceleration. Hence, we use sine.

Moving on to the force of friction, we will use the FUN equation:

Now we need an expression for the normal force, which is the gravitational force multiplied by cos:

Once again, if you're unsure why we used cosine, think about the situation practically. As the hill gets flatter (angle decreased), the normal force grows, and cos is the function that gets larger as the angle gets smaller. Now substituting this into the FUN equation, we get:

Now combining our two forces (remember friction will be subtracted since it is in the opposite direction of movement):

Eliminating mass from the expression:

And rearranging for the coefficient of friction:

We have each of these values, so we can solve:

This will be a two step problem. We need to calculate the coefficient of friction and height of hill. These will be calculated separately. So first, stop and think about which one we can calculate with the information we're given, and which one is dependent on the other.

From the problem statement, we can directly calculate the coefficient of kinetic friction using the velocity of the sledder at the bottom of the hill and the distance it takes for the sledder to come to rest. Since there is no height change, we know that all kinetic energy is dissipated as friction. Therefore, we can say:

Expanding these terms:

Eliminating mass:

Rearranging for the coefficient of friction:

Plugging in the values given in the problem statement:

Now that we have our coefficient of friction, we can move on calculating the height of the hill. For this we can use the expression for conservation of energy:

The sledder is at rest at the top of the hill, 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:

Expanding each term one at a time from left to right:

We need to substitute expression for the normal force and distance that the frictional force is applied. For the normal force we get:

And for distance, we begin with the sine function:

Rearranging for d, we get:

Substituting these into the expression for work, we get:

Cleaning up the equation a bit:

   (4)

Now substituting the expanded expressions back into the first expression, we get:

We can eliminate mass to get:

Now we need to rearrange for height:

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