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A 3.84 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=14.6N at an angle theta=23.0degrees above the horizontal, as shown.

1. What is the magnitude of the acceleration of the block when the force is applied?

2. What is the horizontal speed of the block 5.70 seconds after it starts moving?

3. What is the magnitude of the normal force acting on the block when the force F is acting on it?

4. If, instead, the floor has a coefficient of kinetic friction �k = 0.04, what is the magnitude of the frictional force on the block when the block is moving?

5. What is the magnitude of the acceleration of the block when friction is being considered?

A block located on a horizontal floor pulled by cord exerting force

Edit
Added Wed, 30 Mar '16
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Solutions
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A) Acceleration = (14.6cos23)/3.84 = 3.5 m/s2

B) Speed at 5.70 sec = 3.5*5.70 = 19.95 m/s

C) Normal force = 14.6sin23 = 5.70 N

D) Frictional force = 5.70*0.04 = 0.23 N

E) Acceleration of friction is considered = (14.6cos23-0.23)/3.84 = 3.44 m/s2

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Added Wed, 30 Mar '16
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