Question

A 77.0 kg woman slides down a

Answer 1

Assuming the woman starts at rest, she descends the slide with acceleration a such that

(20.3 m/s)² = 2 a (42.6 m)   →   a ≈ 4.84 m/s²

which points parallel to the slide.

The only forces acting on her, parallel to the slide, are

• the parallel component of her weight, w (//)

• friction, f, opposing her descent and pointing up the slide

Take the downward sliding direction to be positive. By Newton's second law, the net force in the parallel direction acting on the woman is

∑ F (//) = w (//) - f = ma

where m = 77.0 kg is the woman's mass.

Solve for f :

mg sin(42.3°) - f = ma

f = m (g sin(42.3°) - a)

f = (77.0 kg) ((9.80 m/s²) sin(42.3°) - 4.84 m/s²) ≈ 135 N

Compute the work W done by friction: multiply the magnitude of the friction by the length of the slide.

W = (135 N) (42.6 m) ≈ 5770 N•m = 5770 J