M*g = 13 N. = Wt. of puck.
M = 13/g = 13/9.8 = 1.33 kg.
Fn = 13*cos37 = 10.4 N.
Fk = u*Fn = 0.08 * 10.4 = 0.831 N.
sin37 = h/3
h = 3*sin37 = 1.81 m.
PE = Mg*h - Fk*d = 13*1.81 - 0.831*3 =
21.04 J.
At bottom of ramp:
KE = PE = 0.5*M*V^2 = 21.04 J.
0.5*1.33*V^2 = 21.04
V^2 = 21.04/0.663 = 31.72
V = 5.63 m/s = The required initial velocity to reach 3 m.
At the State Fair you see people trying to win a prize at a game booth. They are sliding a metal disk shaped like a puck up a wooden ramp so that it stops in a marked zone near the top of the ramp before sliding back down. You estimate that you can slide the 'puck' at v=3m/s , but would that win the game? Find the distance d that would reach. The two boundaries of the zone appear to be at 3 and 3.2 meters from the bottom of the ramp where you release the 'puck'. The ramp appears to be inclined at 37° from the horizontal. You happen to remember that between steel and wood, the coefficients of static and kinetic friction are 0.1 and 0.08 respectively. The weight of the 'puck' is about W=13N .
1 answer
0 answers