Skip to main content
Ch 06: Dynamics I: Motion Along a Line
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 6, Problem 51a

A 5.0 kg wooden sled is launched up a 25° snow-covered slope with an initial speed of 10 m/s. What vertical height does the sled reach above its starting point?

Verified step by step guidance
1
Step 1: Identify the forces acting on the sled. The sled is moving up the slope, so it is subject to gravitational force, which has a component acting down the slope. The initial kinetic energy of the sled will be converted into potential energy as it climbs the slope.
Step 2: Use the conservation of energy principle. The initial kinetic energy of the sled is given by \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the sled and \( v \) is its initial velocity. The potential energy at the maximum height is given by \( PE = m g h \), where \( g \) is the acceleration due to gravity and \( h \) is the vertical height.
Step 3: Set up the energy conservation equation. At the maximum height, all the initial kinetic energy is converted into potential energy: \( \frac{1}{2} m v^2 = m g h \). Simplify this equation to solve for \( h \): \( h = \frac{v^2}{2 g} \).
Step 4: Account for the slope angle. The vertical height \( h \) is related to the distance traveled along the slope \( d \) by the angle \( \theta \): \( h = d \sin \theta \). Use trigonometry to ensure the vertical height is calculated correctly.
Step 5: Combine the equations. Substitute the values for \( m = 5.0 \; \text{kg} \), \( v = 10 \; \text{m/s} \), \( g = 9.8 \; \text{m/s}^2 \), and \( \theta = 25° \) into the equations to calculate the vertical height \( h \). Ensure units are consistent throughout the calculation.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
12m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 0.5 * m * v², where m is mass and v is velocity. In this scenario, the sled's initial kinetic energy will be converted into potential energy as it moves up the slope.
Recommended video:
Guided course
06:07
Intro to Rotational Kinetic Energy

Potential Energy

Potential energy is the energy stored in an object due to its position in a gravitational field, given by the formula PE = m * g * h, where m is mass, g is the acceleration due to gravity, and h is height. As the sled ascends the slope, its kinetic energy is transformed into potential energy until it reaches its maximum height.
Recommended video:
Guided course
07:24
Potential Energy Graphs

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this problem, the total mechanical energy (kinetic plus potential) of the sled remains constant as it moves up the slope, allowing us to equate the initial kinetic energy to the potential energy at the maximum height.
Recommended video:
Guided course
06:24
Conservation Of Mechanical Energy
Related Practice
Textbook Question

The 2.0 kg wood box in FIGURE P6.58 slides down a vertical wood wall while you push on it at a 45° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?

5
views
Textbook Question

Sam, whose mass is 75 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of 200 N and a coefficient of kinetic friction on snow of 0.10. Unfortunately, the skis run out of fuel after only 10 s. What is Sam's top speed?

Textbook Question

A 1.0 kg wood block is pressed against a vertical wood wall by the 12 N force shown in FIGURE P6.57. If the block is initially at rest, will it move upward, move downward, or stay at rest?

Textbook Question

A baggage handler drops your 10 kg suitcase onto a conveyor belt running at 2.0 m/s. The materials are such that μs = 0.50 and μk = 0.30. How far is your suitcase dragged before it is riding smoothly on the belt?

Textbook Question

A large box of mass M is moving on a horizontal surface at speed v₀. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively. Find an expression for the shortest distance dmin in which the large box can stop without the small box slipping.

Textbook Question

Sam, whose mass is 75 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of 200 N and a coefficient of kinetic friction on snow of 0.10. Unfortunately, the skis run out of fuel after only 10 s. How far has Sam traveled when he finally coasts to a stop?