Work on a system with friction

[kraaaaamos]

Homework Statement

a. How much work must you do to push a 10 kg block of steel across a steel table at a steady speed of 1.0 m/s for 3 s? The coefficient of kinetic friction for steel on steel is 0.60.

b. What is your power output while doing so?

Homework Equations

W = Fd (where d = distance)
F = (fric. coeff)N (where N=normal force)
Normal force in this case = weight
w = mg

P = Fv

The Attempt at a Solution

a) w = mg
= 10(-9.8)
= -98

N = 98

F = (fric. coeff)(N)
= (0.6)(98)
= 58.8

W = Fd d = 3m
= (58.8)(3)
= 176.4 J

b)
P=Fv
= (58.8)(1)
=58.8W

 

[Bill Foster]

You got it right. What is your question?

 

[ogb p]

I would like to clarify that the work done in this scenario is actually negative (-176.4 J) since the force of friction is acting in the opposite direction of the displacement. This means that the system loses energy due to the work done by friction.

Additionally, the power output calculated in part b is the power required to maintain the constant speed of the block, not the power output of the person pushing the block. The power output of the person would also need to account for the work done to overcome the force of friction.