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

[chawki]

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)
F_f=0.25*94*9.8
F_f=230.3 N
or should we use the second law of Newton..but then we don't have acceleration

 

[tiny-tim]

hi chawki!

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

yes we do …

it's zero! ​

 

[chawki]

okok i always get a smile when i see the yellow fish replying

so, by applying Newton second law and by projection on xx axis, we get:
F-F_f=m*0
F=F_f
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

 

[tiny-tim]

yup!

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

 

[supratim1]

yes you are right...

 

[chawki]

yup!

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

That's a very good point, it will avoid calculating W if they didn't ask for the work

 

[chawki]

yup!

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

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