How Do You Calculate Final Speed from Work Done on an Object?

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Homework Help Overview

The discussion revolves around calculating the final speed of a crate subjected to a horizontal force on a frictionless surface. The original poster presents data related to the force applied, distance moved, and mass of the crate, seeking to determine the final speed after work is done on the object.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore two main approaches: one using the constant acceleration equation and the other relating work done to kinetic energy. Questions arise regarding the conversion of work into energy and the nature of energy associated with motion.

Discussion Status

Participants have provided different methods for calculating the final speed, with some exploring the relationship between work and kinetic energy. There is an ongoing exploration of concepts related to energy transformation without a clear consensus on the preferred method.

Contextual Notes

The original poster's inquiry is constrained by the need to apply specific formulas and concepts from physics, such as Newton's laws and energy principles, while also navigating the assumptions about the system being frictionless.

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Finding Final Speed Using Work

A person pushes with a horizonal 25N force over a distance of 1.2m on a 4.5kg crate, intially at rest on a frictionless surface. What is the final speed of the crate?

heres my data list..

F=25N
delta d=1.2m
m=4.5kg
v1=0
v2=?
theta=0 degrees

i used the W=fxdeltadxcos(theta) formula to find out the work done was 30J.. but once i have work i don't no what to find next to find the final speed..
 
Last edited:
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There are two ways of approach. One is using the constant acceleration equation

(v2)^2=(v1)^2+2as

where a is the acceleration of the crate and s is the displacement caused by the acceleration, i.e.

v1 = 0, a = F/m (Newton's Second Law) = 25/4.5 = 50/9, s = 1.2, so

v2 = sqrt (2.4*50/9) = sqrt (120/9 ) = sqrt (120)/3

The other is to find the work done by the force = F.d = 25*1.2 = 30J
and then say this is the additional kinetic energy given to the crate. Since the crate was at rest, we can say

0.5*m*(v2)^2 = 30
v2 = sqrt ( 60/4.5 ) = sqrt (120/9) = sqrt (120)/3
 
What is the work done by the force converted into? (If it is moving what kind of energy does it have?)
 
Energy by virtue of its motion is called kinetic energy, and kinetic energy is given by the formula K.E. = 0.5mv^2 where m is the mass of the body and v is the velocity.
 

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