Question

Steve the daredevil drives his motorcycle along a level track at 20 m/s towards a vertical loop, 14 min meter. Steve and his motorcycle have a combined mass of 150 kg, and he coasts around the loop under only the influence of gravity. (Ignore friction ans air resistenace for a and b).

Physics 220, Winter 2019 Name Exam 2A Long Problems (Show your work for full credit) 2 problems 12 points each 2. Steve the daredevil drives his motorcycle along a level track at 20 m/s towards a vertical loop, 14 min diameter. Steve and his motorcycle have a combined mass of 150 kg and he coasts around the loop under only the influence of gravity (ignore friction and air resistance for a and by a) How fast is Steve going at the top of the loop? 10 2014 (98) =8 29 mis ISO Viborg ( b) What is the Normal force on Steve (and his motorcycle) at the top of the loop (magnitude and direction)? W = 150 kg W=isoka U-isoka c) Steve hits the brakes after reaching the bottom of the loop and skids to a halt in distance d. How much work is done by fiction? W= Falcoso (20)(27)(cos(90))= 0 F F (2X(21)(sin(90))=420 d) If the coefficient of kinetic friction is 0.70 what is d? W=(:70)(d) (coso) d=alm W=(.70)(21)(cos(1)

Answer 1

2

a)

v_b = speed at the bottom = 20 m/s

v_t = speed at the top = ?

r = radius of the loop = diameter/2 = 14/2 = 7 m

h = height gained at the top = 14 m

Using conservation of energy

Total energy at the top = Total energy at the bottom

(0.5) m v_t² + m gh = (0.5) m v_b²

(0.5) v_t² + gh = (0.5) v_b²

(0.5) v_t² + (9.8 x 14) = (0.5) (20)²

v_t = 11.2 m/s

b)

m = mass of steve and motorbike combined = 150 kg

N = Normal force acting in down direction towards the center of the loop = ?

F_g = force of gravity acting in down direction towards the center of the loop = mg = 150 x 9.8 = 1470 N

Force equation for the motion of steve at the top is given as

N + F_g = mv_t²/r

N + 1470 = (150) (11.2)²/7

N = 1218 N

c)

W = work done by friction

v_i = initial speed before applying the brakes = 20 m/s

v_f = final speed after coming to stop = 0 m/s

Work done by friction is given as

W = (0.5) m (v_f² - v_i²)

W = (0.5) (150) ((0)² - (20)²)

W = - 30000 J

d)

d = stopping distance after applying the brakes = ?

$\mu$ = Coefficient of kinetic friction = 0.70

Kinetic frictional force is given as

f = $\mu$ mg

f = (0.70) (150) (9.8)

f = 1029 N

$\theta$ = angle between the frictional force and displacement = 180

Work done by frictional force is also given as

W = f d Cos$\theta$

- 30000 = (1029) d Cos180

d = 29.2 m