Pushing a 94kg Crate: Calculating Force, Work & Power

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Homework Help Overview

The discussion revolves around a physics problem involving a 94-kg crate being pushed across a warehouse floor. The problem requires calculating the force needed to move the crate, the work done during the movement, and the power involved, taking into account the coefficient of kinetic friction.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of Newton's second law and the relationship between force, friction, and acceleration. There is discussion about whether acceleration is relevant in this context, given that the crate is moving at a constant speed.

Discussion Status

Some participants have provided calculations for force, work, and power, while others suggest alternative approaches to streamline the process. There is recognition of the potential to avoid certain calculations if not explicitly required by the problem.

Contextual Notes

Participants note the absence of time in the initial problem statement, which leads to discussions about deriving time from distance and speed. The conversation reflects on the assumptions made regarding the motion of the crate and the implications of constant speed on the calculations.

chawki
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Homework Statement


A heavy, 94-kg crate is pushed a distance of 5.2 m at a speed of 0.40 m/s on the floor of a
warehouse. The coefficient of the kinetic friction is 0.25.

Homework Equations


a) Determine the magnitude of the horizontal force needed to move the crate.
b) Find out the work done in moving the crate.
c) What is the power involved in moving the crate?

The Attempt at a Solution


a)
Ff=0.25*94*9.8
Ff=230.3 N
or should we use the second law of Newton..but then we don't have acceleration :bugeye:
 
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hi chawki! :smile:
chawki said:
A heavy, 94-kg crate is pushed a distance of 5.2 m at a speed of 0.40 m/s

or should we use the second law of Newton..but then we don't have acceleration :bugeye:

yes we do …

it's zero! :biggrin:
 
okok i always get a smile when i see the yellow fish replying :smile:

so, by applying Newton second law and by projection on xx axis, we get:
F-Ff=m*0
F=Ff
F=0.25*94*9.8
F=230.3 N

b) the work done:
W=F*d
W=230.3*5.2
W=1197.56 J

c) Power involved in moving the crate:
P=W/t
we don't have t, but we can get it
S=d/t --->t=d/S =5.2/0.4
t=13s

P=W/t
P=1197.56/13
P=92.12 Watt
 
yup! :biggrin:

except slightly quicker would have been to use power = force.speed :wink:
 
yes you are right...
 
tiny-tim said:
yup! :biggrin:

except slightly quicker would have been to use power = force.speed :wink:

That's a very good point, it will avoid calculating W if they didn't ask for the work
 
tiny-tim said:
yup! :biggrin:

except slightly quicker would have been to use power = force.speed :wink:

That's a very good point, it will avoid calculation W and t, if they don't ask for the work
 

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