Question

A worker pushed a 27 kg block 9.2 m along a level floor at a constant speed with force directed 32° below the horizontal. If the coefficient of kinetic friction between block andfloor was 0.20, calculatea.the work done by the worker's force.b.the increase in thermal energy of the block-floor system.

Answer 1

First, let's draw a diagram of the problem:

Since the force applied is not parallel to the direction of the movement, let's calculate the horizontal and vertical components of the force:

`\begin{gathered} F_x=F\cdot\cos32°\\ \\ F_x=0.848F\\ \\ \\ \\ F_y=F\cdot\sin32°\\ \\ F_y=0.53F \end{gathered}`

Now, let's calculate the friction force:

`\begin{gathered} F_(friction)=F_(normal)\cdot\mu\\ \\ F_(friction)=(m\cdot g+F_y)\cdot\mu\\ \\ F_(friction)=(27\cdot9.8+0.53F)\cdot0.2\\ \\ F_(friction)=52.92+0.106F\text{ N} \end{gathered}`

If the block is moving at constant speed, the horizontal component of the applied force is equal to the friction force:

`\begin{gathered} F_(friction)=F_x\\ \\ 52.92+0.106F=0.848F\\ \\ 0.742F=52.92\\ \\ F=71.32\text{ N} \end{gathered}`

a.

The work done by the worker force is given by:

(the vertical component of this force does no work, since there is no vertical movement. All the work is done by the horizontal component)

`\begin{gathered} W=F_x\cdot d\\ \\ W=71.32\cdot0.848\cdot9.2\\ \\ W=556.41\text{ J} \end{gathered}`

b.

The increase in thermal energy is given by the work done by the friction force.

Since the friction force is equal to the horizontal component of the applied force, the work done by them has the same magnitude, so the increase in thermal energy is 556.41 J.