Question

The figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angle ? = 20

Answer 1

a) Friction force = ?s . Normal force

Normal force = mg - ( F sin 20) Thats the upward component of the pulling force.

So friction force = 0.62 (mg - 0.5.mg.sin20) = 0.514.mg

This force is less than the horizontal pulling force so the block will accelerate. Once it starts moving the frictional force opposing the motion is determined by the kinetic coeff.
Accelerating force = 0.5 * mg * cos20 - 0.500(mg - .5*mg*sin20)

F=0.0553mg=ma

a=0.542

b) Max friction force this time =
0.400(mg - 0.50*mg*0.342)
=.3316 mg

This force is less than the horizontal pulling force so the block will accelerate. Once it starts moving the frictional force opposing the motion is determined by the kinetic coeff.
Accelerating force = 0.5 . mg . cos20 - 0.330(mg - .5mgsin20)

= 0.470 mg - 0.274 mg = 0.196 mg
F = m .a

0.196 mg = m . a

m's cancel so a (the acceleration you want) = .196g
= 1.922 m/s/s (if you take g as 9.81 m/s/s)
= 1.922 m/s/s