A120 kg santa claus slides down a 9 m chimney. he drops from his sleigh 2 m above the top of the chimney. it’s a tight fit, but he slides the whole way. he just comes to rest as he reaches the hearth. what is the force of friction on santa from the chimney?
a) 750 n
b) 2000 n
c) 2500 n
d) 3200 n
e) 1500 n

E) 1500 N

Explanation:

There are two ways to solve this: energy equations or kinematics.

First I'll use energy equations.

All of Santa's energy is converted to work by friction.

Initial energy = final energy + work

PE = W

mgh = Fd

(120 kg) (9.8 m/s²) (9 m + 2 m) = F (9 m)

F = 1437 N

Using kinematics, the velocity Santa reaches when he reaches the chimney is:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2 (-9.8 m/s²) (9 m - 11 m)

v = -6.26 m/s

Then he starts decelerating down the chimney.  Finding the acceleration:

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (-6.26 m/s)² + 2 a (0 m - 9 m)

a = 2.18 m/s²

Sum of the forces acting on Santa:

∑F = ma

F - W = ma

F = W + ma

F = mg + ma

F = m (g + a)

F = (120 kg) (9.8 m/s² + 2.18 m/s²)

F = 1437 N

Rounded to 2 sig-figs, that's 1400 N, which isn't one of the choices.  But if we use 10 m/s² for g instead of 9.8 m/s², we get F = 1467 N, which rounds to 1500 N.

E) 1500 N

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