An 10 kg mass has an applied force of magnitude 69N acting on it pushing it to move in the positive x-direction. The mass is on a flat surface that has a coefficient of kinetic (Uk) friction of 0.5.with respect to the object. What is the magnitude of the acceleration of the object?

The net force on the mass in the vertical direction is

∑ F = n - w = 0

and the net horizontal force is

∑ F = A - f = ma

where

n = magnitude of the normal force

w = mg = weight of the mass

m = 10 kg, mass of the … uh, mass

g = 9.8 m/s², mag. of acceleration due to gravity

A = 69 N, mag. of the applied force

f = mag. of kinetic friction

From the first equation, we get

n = mg = (10 kg) (9.8 m/s²) = 98 N

Then the friction force has magnitude

f = µ n = 0.5 (98 N) = 49 N

Solve the mass's acceleration in the second equation:

69 N - 49 N = (10 kg) a

a = (20 N) / (10 kg) = 2.0 m/s²

in the positive x direction.

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.