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Two boxes connected by a light horizontal rope are on a horizontal surface. The coefficient of kinetic fric-tion between each box and the surface is ilk = 0.30. Box B has mass 5.00 kg, and box A has mass m. A force F with magnitude 40.0 N and direction 53.1° above the horizontal is applied to the 5.00-kg box, and both boxes move to the right with a = 1.50 m/s2. A) What is the tension T in the rope that connects the boxes? B) What is m?

Respuesta :

Answer:

(A). The tension in the rope that connects the boxes is 10.50 N.

(B). The value of m is 7 kg.

Explanation:

Given that,

Mass of box B = 5.00 kg

Mass of box A = m

Force = 40.0 N

Direction= 53.1°

Acceleration = 1.50 m/s²

Coefficient of kinetic friction = 0.30

(A). We need to calculate the tension in the rope that connects the boxes

Using balance equation

[tex]T=ma+m\cos\theta[/tex]

Put the value into the formula

[tex]T=5\times1.50+5.00\cos53.1[/tex]

[tex]T=10.50\ N[/tex]

(B). We need to calculate the value of m

Using formula of tension

[tex]T=ma[/tex]

[tex]m=\dfrac{T}{a}[/tex]

Put the value into the formula

[tex]m=\dfrac{10.50}{1.50}[/tex]

[tex]m=7\ kg[/tex]

Hence, (A). The tension in the rope that connects the boxes is 10.50 N.

(B). The value of m is 7 kg.

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