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A cart loaded with bricks has a total mass of 20.4 kg and is pulled at constant speed by a rope. The rope is inclined at 26.1 â—¦ above the horizontal and the cart moves 20.1 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0.6 . The acceleration of gravity is 9.8 m/s 2 . What is the normal force exerted on the cart by the floor? Answer in units of N.

Respuesta :

Answer:

Normal force, N = 154.5 N

Explanation:

Given that,

Total mass of the cart, m = 20.4 kg

Angle of inclination, [tex]\theta=26.1^{\circ}[/tex]

Distance moved, d = 20.1 m

The coefficient of kinetic friction between ground and cart is 0.6

The value of acceleration due to gravity, [tex]g=9.8\ m/s^2[/tex]

Let N is the normal force exerted on the cart by the floor. It is given by :

[tex]N=mg-T\ sin\theta[/tex]

[tex]T\ cos\theta=\mu N[/tex]

[tex]T\ cos\theta=\mu (mg-T\ sin\theta)[/tex]

[tex]T\ cos(26.1)=0.6 (20.4\times 9.8-T\ sin(26.1))[/tex]

T = 103.23 N

So, the normal force is :

[tex]N=20.4\times 9.8-103.23\ sin(26.1)[/tex]

N = 154.5 N

So, the normal force exerted on the cart by the floor is 154.5 N. Hence, this is the required solution.

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