# Force problem (not for the faint of heart)

• notallthere
In summary, Bob slipped and fell while trying to pull a sled up a 50 degree hill, and the sled started sliding down the hill. The sled traveled 30 meters down the hill and then 50 meters on a wide open horizontal surface before coming to a stop. The coefficient of kinetic friction between the sled and the ground was twice the coefficient of static friction, and the sled would not have moved at all if the coefficient of static friction was twice the coefficient of kinetic friction.
notallthere

## Homework Statement

Bob is pulling a sled up a 50 degree hill (meaning the incline is at an angle of

fifty degrees above the horizontal). Sitting on the sled is Bob's niece (unknown

mass). After pulling the sled 30.0 meters up the incline, Bob slips and falls

releasing the rope attached to the sled. The sled (starting from rest) starts sliding down

the incline. After traveling the 30.0 meters down the hill, the sled travels 50.0 meters on

a wide open horizontal surface before finally coming to a stop. Assuming the coefficient

of kinetic friction is the same throughout the entire problem (hill and level ground), find

the coefficient of kinetic friction. If the coefficient of static friction between the sled and

the ground was twice the coefficient of kinetic friction, would the sled move at all?

Hint #1: Treat this as two different parts with two different coordinate systems

(Part 1- on the hill Part 2- on the level ground)

Hint #2: The speed that the sled reaches at the bottom of the hill is the same speed the

sled starts with when traveling on the level ground.

## Homework Equations

F=ma Ff=μFn Fgx=mgsinθ Fgy=mgcosθ Δx=viΔt+(1/2)aΔt^2 vf^2=vi^2+2aΔx Δx=vfΔt-(1/2)aΔt^2 vf=vi+aΔt Δx=(1/2)(Vi+Vf)Δt
and any other equation that could be used to solve this problem
(kinematic and force equations)

## The Attempt at a Solution

I have no clue on how to solve this problem, and any help would be greatly appreciated

Welcome to the PF.

You *must* show some effort in working toward the solution before we can offer any tutorial help. That's in the PF Rules that you agreed to when you joined (see Site Info at the top of the page).

Start problems like this with a free body diagram (FBD), and list all of the forces acting on the FBD. You will use two -- one for on the slope, and one for on the flat.

The first thing to check with a dynamics problem is whether you can use considerations of energy. If you can, this is often quicker than developing the force and acceleration equations. A good test is whether you are told or asked for the time elapsed. If not, energy is probably the right way.

Suppose the coefficient of kinetic friction is μk. Can you figure out the magnitude of the frictional force when sliding down the hill? What work is done overcoming that? What does that tell you about the KE at the bottom of the hill?

Bob did well to pull a sled up a 50 degree slope! That would be a grade I winter climb. As the title of the post says: not for the faint of heart!

.

I would first start by identifying all the given information and organizing it in a way that makes it easier to understand and solve the problem. From the given information, I can determine that the sled is initially at rest and is pulled up a 50 degree incline for a distance of 30.0 meters. Bob then slips and falls, releasing the sled, which then slides down the incline for a distance of 30.0 meters before reaching a flat surface and traveling an additional 50.0 meters before coming to a stop. The coefficient of kinetic friction is the same throughout the entire problem, and the coefficient of static friction is twice the coefficient of kinetic friction.

I would then draw a free body diagram for each part of the problem - one for when the sled is on the incline and another for when it is on the level ground. This will help me visualize the forces acting on the sled and determine which equations to use.

For the first part of the problem, on the incline, the only forces acting on the sled are the force of gravity (Fg) and the force of friction (Ff). The normal force (Fn) is equal to the component of the force of gravity perpendicular to the incline (Fgy), which can be calculated using Fgy = mgcosθ. The force of friction can be calculated using Ff = μFn, where μ is the coefficient of kinetic friction. Using F=ma, I can then determine the acceleration of the sled on the incline.

For the second part of the problem, on the level ground, the forces acting on the sled are the force of friction (Ff) and the force of gravity (Fg). The normal force (Fn) is equal to the force of gravity (Fg) since the sled is on a flat surface. Using F=ma, I can determine the acceleration of the sled on the level ground.

To find the coefficient of kinetic friction, I would use the kinematic equation vf^2 = vi^2 + 2aΔx, where vf is the final velocity, vi is the initial velocity (which is 0 in this case), a is the acceleration, and Δx is the distance traveled. I would use this equation for both parts of the problem and set the final velocity for the first part (on the incline) equal to the initial velocity for the second part (on the level ground). This will

## What is force?

Force is a physical quantity that can cause an object to accelerate or change its state of motion. It is measured in units of Newtons (N) and is represented by the symbol F.

## What is Newton's first law of motion?

Newtons's first law of motion, also known as the law of inertia, states that an object at rest will remain at rest and an object in motion will continue to move in a straight line at a constant velocity unless acted upon by an external force.

## How do you calculate force?

Force can be calculated by multiplying an object's mass (m) by its acceleration (a). This is represented by the equation F=ma.

## What is the difference between mass and weight?

Mass is a measure of the amount of matter in an object, while weight is a measure of the force exerted on an object by gravity. Mass is constant, while weight can vary depending on the strength of the gravitational force.

## What is friction and how does it affect force?

Friction is a force that resists motion between two surfaces in contact. It acts in the opposite direction of an object's motion and can decrease the magnitude of the force applied to an object.

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