1.A 70-kg diver jumps off a 9.7-m tower. (a) Find the diver\'s velocity when he
ID: 1694648 • Letter: 1
Question
1.A 70-kg diver jumps off a 9.7-m tower.(a) Find the diver's velocity when he hits the water. ____m/s
(b) The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water. __N
2.A 4530-kg helicopter accelerates upward at 2.1 m/s2. What lift force is exerted by the air on the propellers?___ N
3.A 1.0-kg mass (mA) and a 6.0-kg mass (mB) are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. Find the acceleration of the larger mass.
___m/s2 (downward)
Explanation / Answer
Mass m = 70 kg Height h = 9.7 m (a). required velocity v = sqrt [ 2gh ] = 13.78 m / s (b). distance moved before stop S = 2 m Final velocity V = 0 Initial velocity u = v = 13.78 m / s From the relation V^ 2- u^ 2= 2aS Acceleration a = [V^ 2-u^ 2] / 2S = - 47.53 m / s^ 2 Required force f = ma = 3327.1 N (2).mass m = 4530 kg upward acceleration a = 2.1 m / s^2 lift force F = m( a+ g) = 53907 N (3). Masses m = 1 kg M = 6 kg For mass M is Mg – T = Ma From this tension T = Mg –Ma -----( 1) For mass m , T –mg = ma T = mg + ma -----( 2) From eq( 1) and ( 2) , Mg-Ma = mg+ma a(m+M) = g(M-m) from this acceleration a = [ (M-m) / (m+M) ] g = 7 m / s^ 2
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