A 5.48 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a...
Question:
A 5.48 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F = 13.6 N at an angle {eq}\theta {/eq} = 16.5{eq}^\circ {/eq} above the horizontal. What is the magnitude of the acceleration of the block when the force is applied?
Net Force and Acceleration:
The sum of every force that acts on an object is called the net force ("net" is another word for "total"). Remember that forces are vectors, so we need to account for their direction when adding them.
{eq}\vec F_{net} = \vec F_1 + \vec F_2 + \vec F_3 + ... {/eq}
The net force tells us how the motion of an object changes over time. Newton's second law states that the net force on an object is equal to its mass times its acceleration (which always points in the same direction as the net force).
{eq}\vec F_{net} = m\vec a {/eq}
Answer and Explanation: 1
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View this answerThe answer is 2.38 m/s{eq}^2 {/eq}.
We can solve this problem using Newton's second law. The net force on the block equals its mass times...
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