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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 ◦ 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.

User Nokheat
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8.6k points

1 Answer

4 votes

Answer:

Normal force, N = 154.5 N

Step-by-step explanation:

Given that,

Total mass of the cart, m = 20.4 kg

Angle of inclination,
\theta=26.1^(\circ)

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,
g=9.8\ m/s^2

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


N=mg-T\ sin\theta


T\ cos\theta=\mu N


T\ cos\theta=\mu (mg-T\ sin\theta)


T\ cos(26.1)=0.6 (20.4* 9.8-T\ sin(26.1))

T = 103.23 N

So, the normal force is :


N=20.4* 9.8-103.23\ sin(26.1)

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.

User Setzer
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8.1k points
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