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a student attaches a rope to a 20.0 kg box of books. he pulls with force of 90.0 n at a angle of 30.0° with the horizontal. the coefficient of kinetic friction between the box and the sidewalk is 0.500. find the acceleration of the box

Respuesta :

the acceleration of the box is 4.75 m/s^2

Answer:

acceleration = 0.12 m/s^2  

Explanation:

Using Second Law of Motion,

F = m*a

  • F is net force (N)
  • m is mass of the box (kg)
  • a is acceleration of the box (m/s^2)

μ = f/Fn

  • μ is coefficient of friction
  • f is the friction force (N)
  • Fn is the nomal force acting on the object (N)  

Fn = m*g – 90*sin(30)

Fn = (20)*(9.81) – 90*sin(30)

Fn = 151.2 N

f = μ*Fn

f = (0.5)*(151.2)

f = 75.6 N

F = 90cos(30) – 75.6

F = 2.34 N

a = F/m

a = 2.34/20  

a = 0.12 m/s^2  

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