# Horizontal force of 50N accelerates a block of mass 3.0 kg up an incline with an A=2.0ms^-2. The angle with horizontal is 30^o. Find a) force of friction between the box and the incline b) the coefficient of kinetic friction between box and the incline?

##### 1 Answer

May 20, 2018

(a) Let us take the positive

The forces acting on the block along the

- Component of applied force
#=50\ cos 30^@# . - the component of weight
#=-mg\ sin30^@# - the force of friction,
#-f_k# .

Therefore, net force equation along the

#sumF_x = 50\ cos30^@ - mg\ sin30^@ - f_k = ma_x#

Inserting given values we get

#50\ sqrt3/2 - 3xx9.81xx 1/2 - f_k = 3xx2#

#=> f_k = 50\ sqrt3/2 - 3xx9.81xx 1/2 -6#

#=> f_k = 22.6\ N#

(b) The forces acting on the block along the

- Component of applied force
#=50\ sin 30^@# . - the component of weight
#=mg\ cos30^@#

#:.N = mg\ cos30^@+50\ sin 30^@#

Inserting various values we get

#N = 3xx9.81xxsqrt3/2+50xx1/2#

#N = 50.49\ N#

Now he equation relating the friction force and the coefficient of kinetic friction is

#f_k = mu_kN#

#=>mu_k = (f_k)/N#

Plugging in known values we get

#mu_k = (22.6)/50.49=0.45#