How Fast Does a Block Travel on a Frictionless Surface After 3.7 Seconds?

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Homework Help Overview

The discussion revolves around a physics problem involving a block on a frictionless surface being pulled by a force at an angle. Participants are exploring how to determine the speed of the block after a specified time interval, focusing on the concepts of force, mass, and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to apply Newton's second law to relate the force acting on the block to its acceleration. There are questions about how to account for the angle of the applied force in the calculations.

Discussion Status

Some participants have identified relevant principles and laws, while others are seeking clarity on how to apply these concepts, particularly regarding the components of the force due to the angle. There is an ongoing exploration of how to calculate acceleration from the given information.

Contextual Notes

Participants are considering the assumption that the vertical component of the force does not lift the block off the surface, which may influence their calculations. There is also a mention of the time frame for the block's motion, which is central to the problem.

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Block on frictionless floor!

A 3.70 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=13.0 N at an angle q= 25.0° above the horizontal, as shown.





What is the speed of the block 3.70 seconds after it starts moving?

I have drawn a picture but have no idea what equation to start with. know you solve for aceeleration, then once you have accel. you just solve for Vfinal
 
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This involves force, mass and a change in speed. What principle of law relates these physcial concepts?
 


Newtons 2nd law
 


so what now? i know the equations but don't konw how to get acceleration cuase of the theta stuff
 


Assuming the vertical component is not sufficient to lift the block off the floor, use the 2nd law with the horizontal component of the applied force.
 

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