Barney walks at a velocity of 1.7 meters/second on an inclined plane, which has an angle of 18.5° with the ground. what is the horizontal component of barney’s velocity?

Explanation:

Conceptual analysis

Since the velocity has a vector character, in the attached figure we define a rectangular coordinate plane to decompose the total velocity (v) that is in the direction θ with the horizontal.

and are the sides of a right triangle that has hypotenuse v.

The total velocity vector is defined as follows:

Formula (1)

Where:

= vcosθ: Formula (2) magnitude of velocity in x direction

= vsinθ: Formula (3) magnitude of velocity in y direction

i: is positive (+) to the right of the x axis and negative (-) to the left of the x axis

j: is positive (+) upward of the x axis and negative (-) downward of the y axis

Known data

v = 1.7 m/s

θ = 18.5°

Problem development

We replace the data in formula (2) to calculate the magnitude of :

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever