Answer to Question #101775 in Mechanics | Relativity for Titomi

Question

A body of mass 5.0 kg is drawn at a constant speed through a distance 6 m along a horizontal
floor by a rope exerting a constant force of magnitude 10N at an angle of 300 with the horizontal. Calculate the work done by friction on the block

Answer 1

Since the movement occurs at a constant speed, the sum of all the forces acting on the block is zero. In the direction of movement of the block, the projection of the rope tension force on the direction of movement is $F_x=F\cdot cos\alpha=10N\cdot cos30\degree=10 \frac{\sqrt 3}{2}N=5\sqrt 3 N$ acts as shown in the figure.

The friction force $F_f$ acts on the block in the opposite direction, and has equal magnitude.

By definition of work done by force we have $A=\vec F \cdot \vec L$ .

Since the friction force and direction of movement are opposite, their product has a negative sign.

$A= - F_f \cdot L=-5 \sqrt3 \cdot 6 Nm=-30 \sqrt3 J=-52J$

Answer: the work done by friction on the block equals -52 J