Work, energy with kinetic friction

Click For Summary
SUMMARY

The discussion focuses on calculating the distance a box slides after descending a frictionless hill and encountering kinetic friction. The initial velocity at the bottom of the hill is determined using the formula v = √(2gh), yielding a result of 11.89 m/s. The work done by kinetic friction is calculated with W = μk * m * g * d, where μk is the coefficient of kinetic friction (0.246). The final distance the box travels before coming to rest is established as 25.8 meters.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy equations
  • Familiarity with Newton's Second Law of Motion
  • Knowledge of the coefficient of kinetic friction and its application
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the derivation and application of the work-energy principle in physics
  • Learn about the effects of different coefficients of friction on motion
  • Explore advanced kinematics, including motion with varying acceleration
  • Investigate real-world applications of friction in engineering and design
USEFUL FOR

Students studying physics, educators teaching mechanics, and engineers involved in motion analysis and friction-related projects.

Bcisewski
Messages
11
Reaction score
0
Can anyone provide some assistance? I know this comes in two parts, one in locating the velocity, which I believe comes down to v=sq root of 2(9.8)(6.34), however the second part is creating havoc. Any suggestions on part II's formula?

A box slides down a frictionless 6.34 m high hill, starting from rest. At the bottom of the hill, the box slides along a level surface where the coefficient of kinetic friction uk = 0.246. How far from the bottom of the hill does the box come to rest? The final answer will be 25.8m
 
Physics news on Phys.org
The First part

mgh = \frac{1}{2}mv^2

The Second Part:

1)

Use the kinetic constant acceleration formula

v^2 = v_{o}^2 + 2a(x-x_{o})

and Newton's 2nd Law on the box where there's friction to find the acceleration

\sum_{i=1}^{n} \vec{F}_{i} = m \vec{a}

2)

Use

\Delta E = W_{f}

\frac{1}{2}mv^2 = \mu mgd
 
Last edited:


Sure, I can provide some assistance with this problem. First, let's break down the problem into smaller steps to make it easier to solve.

Step 1: Finding the velocity at the bottom of the hill
To find the velocity at the bottom of the hill, we can use the equation v = √(2gh), where g is the acceleration due to gravity (9.8 m/s²) and h is the height of the hill (6.34 m). Plugging in the values, we get v = √(2*9.8*6.34) = 11.89 m/s.

Step 2: Finding the work done by kinetic friction
The work done by kinetic friction can be calculated using the equation W = μk * m * g * d, where μk is the coefficient of kinetic friction, m is the mass of the box, g is the acceleration due to gravity, and d is the distance traveled. Since we are given the values for μk, g, and d, we just need to find the mass of the box to calculate the work done.

Step 3: Finding the mass of the box
To find the mass of the box, we can use the equation F = ma, where F is the force of gravity and a is the acceleration due to gravity. Since the box is sliding down the hill, the force of gravity is equal to the force of kinetic friction. Therefore, we can set the two equations equal to each other and solve for the mass. This gives us m = μk * m * g. Plugging in the values, we get m = 0.246 * m * 9.8. Solving for m, we get m = 1.22 kg.

Step 4: Finding the distance traveled along the level surface
Now that we have the mass of the box, we can use the equation W = μk * m * g * d to find the distance traveled along the level surface. We know that the work done by kinetic friction is equal to the change in kinetic energy, so we can set W = 1/2 * m * v^2 (kinetic energy equation). This gives us μk * m * g * d = 1/2 * m * v^2. Cancelling out the mass on both sides, we get μk * g * d = 1/2 * v^2. Pl
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
2K
Replies
18
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
48
Views
8K
Replies
7
Views
1K
Replies
2
Views
13K
  • · Replies 8 ·
Replies
8
Views
4K