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A 20 kg wooden crate is dragged 12 m across a rough level floor by a rope. The force of tension in the rope is 50 N at an angle of 25° above the horizontal and the crate is moving at a constant speed of 1.50 m/s. Find the magnitude of the normal force [N].

User Marichu
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1 Answer

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To solve this problem, we can use the equations of motion and Newton's second law of motion. The equation of motion for an object moving at constant speed is:

x = x0 + v0t + (1/2)at^2

where x is the final position, x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration.

In this case, the initial position is 0 m (since the crate starts at the beginning of the 12 m distance), the initial velocity is 0 m/s (since the crate starts at rest), and the time is 12 s (since the crate moves at a constant speed of 1.50 m/s).

We can use these values to solve for the acceleration:

12 m = (1/2)(a)(12 s)^2

a = 0.5 m/s^2

Now that we know the acceleration, we can use Newton's second law of motion to find the normal force:

F = ma

N = (20 kg)(0.5 m/s^2)

N = 10 N

This is the magnitude of the normal force on the crate. Note that this is the normal force exerted by the floor on the crate, not the force of tension in the rope. To find the force of tension in the rope, we would need to consider the angle of the rope and the direction of the force.

User Bhagyas
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