Question

A worker pushed a 30.0 kg block 6.50 m along a level floor at constant speed with a force directed 30.0° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.190, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

Answer 1

The work done by worker's force is same as thermal energy 327.2 J

Step-by-step explanation:

Given :

Mass of block
`m = 30` Kg

Displacement
`d = 6.50` m

Angle
`\theta =` 30°

Coefficient of kinetic friction
`\mu _{} = 0.190`

(A)

The work done by the worker's

`W = Fd\cos \theta`

Here force is given by,

`F \cos \theta = \mu F_(n)`

Where
`F_(n) = mg + F \sin \theta`

`F \cos \theta = \mu (mg + F\sin \theta )`

`F \cos 30 + 0.19( F\sin 30 )= \mu m g`

`0.866 F + 0.095 F = 0.19 * 30 * 9.8`

`F = (55.86)/(0.961)`

`F = 58.13` N

So work done by,

`W = 58.13 * 6.50 * \cos 30`

`W = 327.2` J

(B)

The thermal energy of the block floor system is,

`E_(th) = f_(k) d`

`E_(th) = 58.13 * 6.50 * \cos 30`

`E_(th) = 327.2 J`

Therefore, the work done by worker's force is same as thermal energy 327.2 J