A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. She slows steadily, then continues on at 5.8 m/s . What is her acceleration on the rough ice?

Recall that

v² - u² = 2 a ∆x

where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled (because the skater moves in only one direction).

So we have

(5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s²

b.

hope this : )

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2