Can someone please help me with these physics problems? I just don’t even know where to start.

#1
for the block of mass 5 kg normal force is given as
[tex]F_n = mg[/tex]
[tex]F_n = 5*9.8 = 49 N[/tex]
friction force is given as
[tex]F_f = \mu F_n[/tex]
[tex]F_f = 0.1*49 = 4.9 N[/tex]
Net force is given as
[tex]F_{net} = ma[/tex]
[tex]F_{net} = 5*2 = 10 N[/tex]
now we know that
[tex]F_{net} = F_{app} - F_f[/tex]
[tex]10 = F_{app} - 4.9[/tex]
[tex]F_{app} = 14.9 N[/tex]
#2
Normal force is given as
[tex]F_n = mg[/tex]
[tex]F_n = 6*9.8[/tex]
[tex]F_n = 58.8 N[/tex]
now we know that
[tex]F_{net} = F_{app} - F_f[/tex]
[tex]F_{net} = 0[/tex]
as object moves with constant velocity
[tex]F_{app} = F_f = 15 N[/tex]
now for coefficient of friction we can use
[tex]F_f = \mu F_n[/tex]
[tex]15 = \mu * 58.8[/tex]
[tex]\mu = 0.255[/tex]
#3
net force upwards is given as
[tex]F = 1.2 * 10^{-4} N[/tex]
mass is given as
[tex]m = 7 * 10^{-5} kg[/tex]
now as per newton's law we can say
[tex]F = ma[/tex]
[tex]1.2 * 10^{-4} = 7 * 10^{-5} * a[/tex]
[tex]a = 1.71 m/s^2[/tex]
#4
As we know that when block is sliding on rough surface
part a)
net force = applied force - frictional force
[tex]F_{net} = F_{app} - F_f[/tex]
[tex]ma = F_{app} - F_f[/tex]
[tex]5*6 = 40 - F_{f}[/tex]
[tex]F_f = 40 - 30 = 10 N[/tex]
part b)
for coefficient of friction we can use
[tex]F_f = \mu F_n[/tex]
[tex]10 = \mu * F_n[/tex]
here normal force is given as
[tex]F_n = mg = 5*9.8 = 49 N[/tex]
now we have
[tex]\mu = \frac{10}{49} = 0.204[/tex]
#5
if an object is initially at rest and moves 20 m in 5 s
so we can use kinematics to find out the acceleration
[tex]d = v_i*t + \frac{1}{2}at^2[/tex]
[tex]20 = 0 + \frac{1}{2}a(5^2)[/tex]
[tex]a = 1.6 m/s^2[/tex]
now net force is given as
[tex]F_{net} = ma[/tex]
[tex]F_{net} = 10*1.6 = 16 N[/tex]
#6
an object travelling with speed 25 m/s comes to stop in 1.5 s
so here acceleration of object is given as
[tex]a = \frac{v_f - v_i}{t}[/tex]
[tex]a = \frac{0 - 25}{1.5} = -16.67 m/s^2[/tex]
now the force is gievn as
[tex]F = ma[/tex]
[tex]F = 5*16.67 = 83.3 N[/tex]