# physics help?

(a) A maple samara or "helicopter" of mass 67 mg falls 3.1 m, starting from rest. The trip to the ground takes 5.0 s. What is the magnitude of the upward force of lift exerted by the air on the samara?

(b) A 62-kg parachutist is falling at 53 m/s when her parachute opens. She falls a further 148 m in 5.0 s. What was the force exerted on the parachutist by the parachute? (Neglect the much smaller force of air resistance on the parachutist herself.)

### 4 Answers

- oubaasLv 73 weeks ago
(a)

A maple samara or "helicopter" of mass 67 mg falls 3.1 m, starting from rest. The trip to the ground takes 5.0 s. What is the magnitude of the upward force of lift exerted by the air on the samara?

acceleration a = 2h/t^2 = 6.2/5^2 = 0.248 m/sec^2

friction force FF = m*(g-a) = 67*10^-6*(9.807-0.248) = 0.64 mN (supposedly constant)

(b) A 62-kg parachutist is falling at 53 m/s when her parachute opens. She falls a further 148 m in 5.0 s. What was the force exerted on the parachutist by the parachute? (Neglect the much smaller force of air resistance on the parachutist herself.)

148 = V*t+a/2*t^2

148 = 53*5+a*25/2

a = (148-265)/12.5 = -9.36 m/sec^2

module of F = m*(g-a) = 62*(9.36+9.807) = 1200 N (two significant figures)

versus of F = opposite to g = up

- FiremanLv 74 weeks ago
(a) Let the acceleration of fall is a m/s^2

=>By s = ut + 1/2at^2

=>3.1 = 0 + 1/2 x a x (5)^2

=>a = 0.25 m/s^2

Thus By F(net) = F(gravity) - F(upward Thrust by air)

=>ma = mg - F(T)

=>F(T) = m x (g-a)

=>F(T) = 67 x 10^-3 x (9.8 - 0.25)

=>F(T) = 0.64 Newton

(b) Let the acceleration of fall is a m/s^2 after the parachute opens,

=>s = ut + 1/2at^2

=>148 = 53 x 5 + 1/2 x (-a) x (5)^2 {a -ve as it is retarding the motion}

=>a = 9.36 m/s^2 (upward)

Thus By F(net) = F(gravity) - F(upward Thrust by air)

=>ma = mg - F(T)

=>F(T) = m x (g-a)

=>F(T) = 62 x (9.8 - 9.36)

=>F(T) = 27.28 Newton

- NCSLv 74 weeks ago
(b) "falls a further 148 m in 5.0 s" makes her average velocity during this time

Vavg = 148m / 5.0s = 29.6 m/s

Vavg = 29.6 m/s = ½(53m/s + V)

where V is her speed at the end of the 5.0 s

V = 6.2 m/s

V² = U² + 2as

6.2² = 53² + 2*a*148m

solves to

a = -9.36 m/s²

so the upward force is

F = m(g - a) = 62kg * (9.81 - -9.36)m/s² = 1189 N

(a) Starting from rest we can rearrange s = ½at² to get

a = 2s / t² = 6.2m / (5.0s)² = 0.248 m/s²

by newton's second,

net force = ma = mg - lift

67E-6kg * 0.248m/s² = 67E-6kg * 9.81m/s² - lift

lift = 6.4E-4 N or 640 µN

Hope this helps!

- oldprofLv 74 weeks ago
(a) A maple samara or "helicopter" of mass m = 67 mg falls h = 3.1 m, starting from rest. The trip to the ground takes T = 5.0 s. What is the magnitude of the upward force of lift exerted by the air on the samara?

From h = 1/2 aT^2 we have a = 2h/T^2 as the acceleration to the ground. And from f = ma = W - Fy we have upward drag force Fy = W - ma = m(g - a) = m(g - 2h/T^2) = .067*(9.81 - 2*3.1/25) = 0.640654 N. ANS.

(b) A M = 62-kg parachutist is falling at V = 53 m/s when her parachute opens. She falls a further s = 148 m in T = 5.0 s. What was the force exerted on the parachutist by the parachute? (Neglect the much smaller force of air resistance on the parachutist herself.)

From s = 1/2 aT^2 we have a = 2s/T^2 as the acceleration while chuted. Assuming the weight of the chutist (with chute) is W = Mg the net force f = W - D = Ma so that D = W - Ma = M(g - a) = 62*(9.81 - 2*148/25) = -126 N upward. ANS.