Question in description (Prism and rectangular parallelepiped)

Question

2.00cm cube of steel is placed in a surface grinding machine, and the vertical feed is set so that 0.1mm of metal is removed from the top of the cube at each cut. How many cuts are needed to reduce the weight of the cube by 8.3g

 

From the technical mathematics with calculus: second canadian edition textbook.

Answer 1

While the density of steel is not specified, let's assume D=8g/cm³ (for calculation simplicity used in this illustration).  Once density value (D) is finalized, the following illustration can be easily revised.

 

Cube dimensions = 2cm x 2cm x 2cm

D=8g/cm³ (assumed)

 

(V)cube=(2cm)³=8cm³.........initial volume of cube.

Cube weight=(W)cube

                  =(V)cube x D

                  =8x8

   ∴(W)cube=64g.................weight of cube before planing.

 

Per prob statement, 8.3g are to be removed, so....

        (W)red=64g-8.3g

                  =55.7g..............final cube weight

 

Now solve for a new cube height using final weight...

        (W)red=(V)red x D

            55.7=2x2xHx8

            55.7=32H

              ∴H=1.74cm

 

         (H)red=2cm-1.74cm

                  =0.26cm

      ∴(H)red=2.6mm............reduced planed thickness

 

With each planing pass being 0.1mm thick.......

      #passes=2.6mm/(0.1mm/pass)

                  =26 passes

 

Lets check our results using our revised dimensions.....

        (W)red=(2cm x 2cm x 1.74cm) x 8g/cm³

                  =6.96cm³ x 8g/cm³

                  =55.68g

 

64g-55.68g=8.32g....check√ (original cube weight minus final cube

                                            weight equals required removed weight

                                            per problem statement)