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forces physics problem - Please Help ASAP

A sky diver of mass 66 kg can slow herself to a constant speed of 99 km/h by orienting her body horizontally, looking straight down with arms and legs extended. In this position, she presents the maximum cross-sectional area and thus maximizes the air-drag force on her.
(a) What is the magnitude of the drag force on the sky diver?

(b) If the drag force is equal to bv2, what is the value of b?
(c) At some instant she quickly flips into a "knife" position, orienting her body vertically with her arms straight down. Suppose this reduces the value of b to 54 percent of the value in Parts (a) and (b). What is her acceleration at the instant she achieves the "knife" position?

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1 Answer

(a) Fairly simple problem here.   If the diver has constant speed then the drag force is equal and opposite to the force of gravity.  Drag = m *g = 66kg * 9.8m/s^2 up = 647 n up

(b)  Be careful regarding the units in part b.  Note the speed is in km/hr.

bv^2 =  mg;   So b = mg/v^2 =   9.8m/s^2 * 66kg/(99km/hr)^2 * (1km/1000m)^2 *(3600s/1hr)^2

b = 0.85 kg/m

(c)  if b is reduced by 0.54  of original avlue then the drag force becomes 0.54 * 647n or 349n.     The net force on the sky diver is then gravity force - drag force = 9.8m/s^2 * 66kg - 349n = 298 n downward.

The acceleration a = F/m 298n/66kg = 4.5m/s^2 downward.