Block pulled on horizontal surface -w/friction-

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Homework Help Overview

The problem involves a box being pulled along a horizontal surface with friction, where the forces acting on it include a pulling force at an angle and the weight of the box. Participants are tasked with finding the kinetic energy and velocity after a certain distance, given specific parameters such as weight, pulling force, angle, and coefficient of friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of normal force and frictional force, with one participant expressing uncertainty about finding the final velocity. Others introduce equations of motion and clarify the variables involved, while some question the definitions of terms used in the equations.

Discussion Status

The discussion is ongoing, with participants providing insights into the equations of motion and the relationships between variables. Clarifications are being sought regarding the meanings of specific terms, and there is an exploration of how work done on the box relates to its motion.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available and the methods they can use to arrive at a solution.

bdragons
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Homework Statement


A box weighing 80N is pulled along a horizontal surface by a force of 15 N which is directed at 20° above the horizontal. If the box started from rest and the coefficient of friction is 0.15, find the kinetic energy and velocity of the box after it traveled a distance of 10 m.

Homework Equations


Normal force= W-Tsinθ
Frictional force=μk x Normal force

The Attempt at a Solution


First I looked for normal force which turns out as 74.87 then frictional force which is 11.23 after that, I don't get how to find velocity. I just need final velocity at this point to get average velocity=velocity then having V, I could solve for K.E.
 
Last edited:
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Force accelerates the box's speed.
It has a speed of 0m/s in the beginning.
[itex]x = x_0 + v_0t + 0.5at^2[/itex]
[itex]x = 0.5at^2[/itex]
On the other hand
[itex]v_{end} = v_0 + at[/itex]
[itex]v_{end} = at[/itex]
Should be all the info you need. Think about it.
 
would just like to clear the following things:
-what does x,x0,v0 stand for? (guessing that 0 there means zero)
-does a stand for acceleration while t is time?
 
They're the basic equations of motion.
x = total distance
x_0 = distance from origin(usually 0 unless origin is set somewhere else than on the system you're investigating)
v_0 = speed the system has when you begin investigating it
a = acceleration
t = time
 
If you know the net force F moves the block, acting in the direction of motion, then the work done on the block moving it distance d is given by F*d. Where does that work end up (how is it expressed in the system)?
 

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