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 .
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.
To determine whether sliding the puck at a speed of 3 m/s would win the game, we need to calculate the distance the puck would reach on the ramp.
First, let's analyze the forces acting on the puck:
1. Weight (W): The force due to gravity pulling the puck downwards. Its magnitude is given as 13 N.
2. Normal force (N): The force exerted by the ramp on the puck in the vertical direction. It is equal in magnitude and opposite in direction to the weight of the puck. So, N = 13 N.
3. Friction force (f): The force opposing the motion of the puck. There are two types of friction forces we need to consider:
a. Static friction: The friction force when the puck is at rest or not sliding. Its magnitude is given by the coefficient of static friction (μs) multiplied by the normal force (N). Here, μs = 0.1, so fs = μs * N.
b. Kinetic friction: The friction force when the puck is sliding. Its magnitude is given by the coefficient of kinetic friction (μk) multiplied by the normal force (N). Here, μk = 0.08, so fk = μk * N.
Since the puck is on the verge of sliding, the static friction force fs just balances the force component of gravity pulling the puck parallel to the ramp. This force component can be calculated using the weight (W) and the angle of inclination (θ) of the ramp.
Method:
1. Calculate the force component of gravity parallel to the ramp: F_parallel = W * sin(θ).
2. If F_parallel is less than or equal to fs, the puck will not slide. In this case, the speed of 3 m/s would not win the game.
3. If F_parallel is greater than fs, the puck will slide. In this case, we need to calculate the distance the puck would reach.
a. Find the net force acting on the puck in the direction of motion: F_net = W * sin(θ) - fk.
b. Apply Newton's second law (F_net = m * a) to find the acceleration of the puck. The mass (m) is not given, but we can assume it to be cancelled out in our calculations.
c. Apply the equations of motion to find the distance (d) using the initial speed (v), acceleration (a), and the equation d = (v^2 - u^2) / (2 * a), where u is the initial velocity.
d. If d is within the range of 3 to 3.2 meters, then sliding the puck at 3 m/s would win the game.
Remember to double-check all the assumptions and calculations to ensure accuracy.