An object with a mass of #5 kg# is on a surface with a kinetic friction coefficient of # 4 #. How much force is necessary to accelerate the object horizontally at # 9 m/s^2#?

1 Answer
Mar 29, 2016

Answer:

#F_x=5(9-.4*10) ~~25N#

Explanation:

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Draw a free body diagram, FBD:
x-axis:
#sumF_x=ma; F_x-F_(mu_k) = ma # Newton Law ===>(1)
Summation of all Forces #=> ma#

y-axis:
#sumF_y=ma; F_w-N = 0; F_w=mg=N # Newton Law
#F_(mu_k)=mu_k *N= mu_k *mg# ===>(2)

Insert (2) into (1):

#F_x - mu_k *mg = ma# Solve for #F_x#
#F_x = ma-mu_k *mg = m(a-mu_kg)#

Now #m=5kg; mu_k = .4; a=9m/s^2#
#F_x=5(9-.4*10) ~~25N#

A word of caution, a kinetic friction, #mu_k = 4# is very large, I suspect you meant to say #.4# which reasonable. I assumed .4. If you want #mu_k = 4# just correct for it.