The acceleration of the block is 4.90 m/s^2.
:
Here are some sample problems and solutions from the 5th edition:
The force acting on the block is:
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.
The direction is:
So, the magnitude of the resultant displacement is 48.2 km, and its direction is 38.3° south of west. The acceleration of the block is 4
R = √(37.5 km)^2 + (-30.3 km)^2 = 48.2 km
Let's break down the displacement into its north and west components:
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 direction is: So, the magnitude of the
a = F / m = (mg * sin(30.0°)) / m = g * sin(30.0°) = 9.80 m/s^2 * 0.500 = 4.90 m/s^2
Using Newton's second law:
θ = tan^(-1)(-30.3 km / 37.5 km) = -38.3° The direction is: So