Respuesta :
Answer:
The frictional force [tex]F_{fri} =[/tex] 6.446 N
The acceleration of the block a = 6.04 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
Mass of the block = 3.9 kg
[tex]\theta = 40[/tex]°
[tex]\mu[/tex] = 0.22
(a). The frictional force is given by
[tex]F_{fri} = \mu R_{N}[/tex]
[tex]R_{N} = mg \cos \theta[/tex]
[tex]R_{N} =[/tex] 3.9 × 9.81 × [tex]\cos 40[/tex]
[tex]R_{N} =[/tex] 29.3 N
Therefore the frictional force
[tex]F_{fri} =[/tex] 0.22 × 29.3
[tex]F_{fri} =[/tex] 6.446 N
(b). Block acceleration is given by
[tex]F_{net} = F - F_{fri}[/tex]
F = 30 N
[tex]F_{fri}[/tex] = 6.446 N
[tex]F_{net}[/tex] = 30 - 6.446
[tex]F_{net}[/tex] = 23.554 N
The net force acting on the block is given by
[tex]F_{net} = ma[/tex]
23.554 = 3.9 × a
a = 6.04 [tex]\frac{m}{s^{2} }[/tex]
This is the acceleration of the block.
