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Here are some sample problems and solutions from the 5th edition:
F = mg * sin(30.0°)
a = F / m = (mg * sin(30.0°)) / m = g * sin(30.0°) = 9.80 m/s^2 * 0.500 = 4.90 m/s^2 : Here are some sample problems and solutions
A car travels 20.0 km due north and then 35.0 km in a direction 60.0° west of north. Find the magnitude and direction of the car's resultant displacement.
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R = √(37.5 km)^2 + (-30.3 km)^2 = 48.2 km : "Principles of Physics: A Calculus-Based Text" by
North component: 20.0 km + 35.0 km * cos(60.0°) = 20.0 km + 17.5 km = 37.5 km West component: -35.0 km * sin(60.0°) = -30.3 km
The acceleration of the block is 4.90 m/s^2.
:
"Principles of Physics: A Calculus-Based Text" by David Halliday, Robert Resnick, and Jearl Walker is a widely used textbook in introductory physics courses. The 5th edition of this book provides a comprehensive introduction to physics, using calculus as a mathematical tool. Here are some key principles and solutions from the text:
Let's break down the displacement into its north and west components:
A 3.00-kg block is pushed up a frictionless ramp that makes an angle of 30.0° with the horizontal. Find the block's acceleration. Find the block's acceleration
Using Newton's second law:
So, the magnitude of the resultant displacement is 48.2 km, and its direction is 38.3° south of west.