Respuesta :

#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]

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