Question

(A) how much work is done when you push a crate horizontally with 100 N across a 10-m factory floor?

(B) if the force of friction in the crate is a steady 70 N, show that the KE gained by the crate is 300 J.
(C) show that 700J is turned into heat.

Answer 1

A) 1000 joules

Step-by-step explanation:

In general work is given by the equation:

`W=\intop_(a)^(b)\overrightarrow{F}\cdot d\overrightarrow{s}` (1)

A) With
`\overrightarrow{s}` the displacement and
`\overrightarrow{F}` the force applied, because the force and the displacement are parallel (the crate is pushed horizontally)
`\overrightarrow{F}\cdot d\overrightarrow{s}` is simply
`F\,ds`, and because the path is a straight line and the force is constant work is:

`W=FS` (2),

`W=(100)(10)=1000\,J`

B) The work-energy theorem says that the total work on a body is equal to the change on kinetic energy:

`W_(tot)=\varDelta K` (3)

The total work on the crate is the work done by the push and plus the work of the friction
`W_(tot)=W + W_(f)` (4) , as (A) because forces are parallel to the displacement
`W= FS` (5) and
`W_(f)=-fS` (6), the due friction always has negative sign because is opposite to the displacement, using (6), (5) and (4) on (3):

`FS-fS=\varDelta K` (3)

`1000-(70)(10)=300\,J`

C) The energy is lost by friction, so the amount of energy turned into heat is the work the friction does:

`Q=fS=(70)(100)=700\,J` (3)