Question

There is an man running a bicycle with weight of 210lb that is moving with a velocity of v=45feet/sec and detects a barrier and applies the brakes a distance D Before the barrier of 20 feet. It hits the first barrier (Barrier 1) of length B1=12 feet with a force of 70lb and then hits the second barrier (barrier 2) that has a length of B2 = 32 feet with a force of 50 lb. The surface has a friction coefficient of 0.60. 1. Determine the velocity the man has before touching barrier 1. 2. Determine the velocity the man has before touching barrier 2. 3. Can the Barriers stop the man? If they can where the man ends? D Before Barrier B1 Distance B 2 Distance Force → Displacement Barrier 1 Barrier 2 B 1 Distance B 2 Distance

Answer 1

weight of (Bicycle + man) = 2101b relocity of Bicycle- 45 feet/sec Ik 12 feet - k 32 feet 1 B2 distance ! O ko t B, B isteme + 20 feet Barrier 2 Barrier 1 1. onee the man applies force F. (Brakes) to stope bicycles. tyres starts skitting and the Fi is friction force acting between tyres and surface OTO F, MIN) - umg = 0.60 x 210x (32.2) = 4057.2 lb auceleration (Deceleration of bikes after Applying brakes) * a u g - 060x22.2 - 19:32 feet (see? 0.60x32.2 = 19.32 feet/see- a = - 19.32 feet/ see

1. bring wook energy theorem between A & B → įm van Love + ungo I m & m - zimga * x210x14 = 3x2104 (0.5%= (050x210x82.2420) VB x 105 = 212625 - 8 1144 WA VB 162 A 6 ) / (35.386 4 m/see) YB = 35.886 m/ see relocity the man was before touching barries 1.

At B man hits the barries with toll force and this acts back! on bicycle due to reaction. so, Again Applying work eneyy theorem - Ilmve 1 mvh + umgx (12) + (70)+(2) >> Š x 2104 (35.386)' x 210 x ve + 0.6x240x32.2 x 12 +70/12 » 1314 77.744 = 105 ve ² + 48686.4 + 840 The = 21.937 peet/ree r relocity of man before touching Barries 2 work encefy between C & D [suppose at a brcycle stops mye mv ung (18) + som -> 4x210x (27.931) = 0 + 0.6X210X32.2 X+ son 81949.976 = 4057. 22 tron x = 19.952 feet => yes the barries can stop the man and he ends at (51.952 feet) k 20 feet 12 feet 19.952feet