In the previous chapter, a weightlifter is shown holding a barbell steady. Class 9

In the previous chapter, a weightlifter is shown holding a barbell steady. Class 9

Pause and Ponder (Page 119)

Question 1.

In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig.). Is she doing any work on the barbell while holding it steady?

Answer:

No. Work done = Force × Displacement. While the weightlifter holds the barbell steady, there is no displacement of the barbell (s = 0). Therefore, work done on the barbell = F × 0 = 0. Even though her muscles use up internal energy (and she feels tired), no scientific work is done on the barbell.

Question 2.

Is the work done by friction on the stack of coins that travels on a rough surface (Fig.) - positive, negative or zero? Class 9

Fig: Force acting on the stack of coins (a) due to the stretched rubber band and friction upon release of rubber band, and (b) only due to friction

Answer:

Negative. The frictional force acts on the stack of coins in a direction opposite to the direction of displacement (friction opposes motion). Since force and displacement are in opposite directions, the work done by friction is negative.