Question

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?

Answer 1

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.