- #1

Argonaut

- 45

- 24

- Homework Statement
- [Sears and Zemansky's University Physics, 13E] 6.3 A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

- Relevant Equations
- Force and Work

I keep running into the issue of getting a different number of significant figures in my solutions than the official answers at the back of the book, so I'm wondering if I'm missing something or if the book is occasionally sloppy about significant figures.

In Unit 1, the book states (p.8.):

And the these are the answers official answers to the exercise above:

Here is my final calculation of a):

$$

F_p = \mu_k m g = (0.25)(30.0~\rm{kg})(9.80~\rm{m/s^2}) = 73.5~\rm{N}

$$

As far as I can tell, each term has three significant figures, unless ##30.0## counts as two significant figures.

And then one of b) or c) has to be a typo as the ##W_p = -W_f##. This is my calculation of ##W_p##:

$$

W_p = F_p s\cos{\phi} = (73.5 ~\rm{N})(4.5 ~\rm{m})(\cos{(0^{\circ})})=330 ~\rm{J}~\text{(to 2 significant places)}

$$

Since the displacement is given to two significant places, ##333 ~\rm{N}## is spurious accuracy, isn't it?

In Unit 1, the book states (p.8.):

Multiplication or Division:

Results may have no more significant figures than the starting number with the fewest significant figures.

And the these are the answers official answers to the exercise above:

6.3

a) 74 N

b) 333 J

c) -330 J

d) 0, 0

e) 0

Here is my final calculation of a):

$$

F_p = \mu_k m g = (0.25)(30.0~\rm{kg})(9.80~\rm{m/s^2}) = 73.5~\rm{N}

$$

As far as I can tell, each term has three significant figures, unless ##30.0## counts as two significant figures.

And then one of b) or c) has to be a typo as the ##W_p = -W_f##. This is my calculation of ##W_p##:

$$

W_p = F_p s\cos{\phi} = (73.5 ~\rm{N})(4.5 ~\rm{m})(\cos{(0^{\circ})})=330 ~\rm{J}~\text{(to 2 significant places)}

$$

Since the displacement is given to two significant places, ##333 ~\rm{N}## is spurious accuracy, isn't it?