Energy Problems of box acceleration

[JassC]

A 29.4 kg box initially at rest is pushed 3.5 m along a rough, horizontal floor with a constant applied horizontal force of 132.853 N. The acceleration of gravity is 9.8 m/s^2 . If the coefficient of friction between box and floor is 0.408, find the work done by the applied force. Answer in units of J.

My understanding that it is asking for the applied force WITHOUT the friction? So just 3.5m * (29.4kg)(9.8m/s^2) = the applied force?

 

[Hootenanny]

A 29.4 kg box initially at rest is pushed 3.5 m along a rough, horizontal floor with a constant applied horizontal force of 132.853 N. The acceleration of gravity is 9.8 m/s^2 . If the coefficient of friction between box and floor is 0.408, find the work done by the applied force. Answer in units of J.

My understanding that it is asking for the applied force WITHOUT the friction? So just 3.5m * (29.4kg)(9.8m/s^2) = the applied force?

You may wish to reconsider your answer there, what is the definition of work done?

 

[PPonte]

Expanding Hootenanny's hint:

W = \Sigma \vec{F} \times \Delta s

and

\Sigma \vec{F} = F - F_{a}

F - constant applied horizontal force of 132.853 N

 

[JassC]

Sorry.

But that didn't help me :(

 

[PPonte]

Sorry.

But that didn't help me :(

OK. No problem.

\Sigma \vec{F} = F - f_k

F - constant applied horizontal force of 132.853 N
f_k - force of kinetic friction

To find the kinetic friction force:

f_k =\mu_k.n
where n is the normal force, and since the box is along an horizontal floor it is equal to the weight of the box.

Then, you calculate \Sigma \vec{F}, using \Sigma \vec{F} = F - f_k

And finally, you compute the work done, using W = \Sigma \vec{F} \times \Delta s, since you now know \Sigma \vec{F} and \Delta s.

I hope this time you understand. If not, we are here to provide more help.

 

[JassC]

Alright I got the correct answer.

Thanks!