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 8.0 ft/sec, but would that win the game? The two boundaries of the zone appear to be at 10 and 10.5 feet 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 2.5 lbs.
(a) What is the algebraic expression for the distance up the ramp you slide the puck in terms of the initial speed with which you slide the puck (V), the angle of the ramp (θ), the coefficients of static and kinetic friction (
μS
and
μK
), the weight of the puck (W), and the gravitational acceleration (g)? [Note: Don't enter an equation like "x=blah". Just enter the "blah" part. All letters are capital except for "g".]
(b) What is the numerical value of the distance you slide the puck up the ramp (make sure to include units and put a space between the number and the units or else the system will not recognize it!!)?
(a) The algebraic expression for the distance up the ramp you slide the puck can be derived using the principles of physics. The net force acting on the puck can be determined by considering the forces acting in the vertical and horizontal directions.
In the vertical direction, the weight of the puck (W) acts downward, and the normal force (N) acts upward. The normal force can be calculated as N = W * cos(θ), where θ is the angle of the ramp.
In the horizontal direction, the force of static friction (F_static) opposes the applied force (F_applied), which is the force exerted by sliding the puck. When the puck loses contact with the ramp and starts sliding back down, the force of kinetic friction (F_kinetic) opposes the motion.
The expression for the distance up the ramp (d) in terms of the given variables is:
d = (V^2 * μS) / (g * (1 - μS * sin(θ)))
where:
- V is the initial speed with which you slide the puck (8.0 ft/s in this case).
- θ is the angle of the ramp (37° in this case).
- μS is the coefficient of static friction between the puck and the ramp (0.1 in this case).
- g is the gravitational acceleration (32.2 ft/s^2).
- d is the distance up the ramp we are trying to determine.
(b) To find the numerical value of the distance you slide the puck up the ramp, we can substitute the given values into the above expression and calculate the result.
Plugging in the values:
V = 8.0 ft/s
θ = 37°
μS = 0.1
g = 32.2 ft/s^2
d = (8.0^2 * 0.1) / (32.2 * (1 - 0.1 * sin(37°)))
Evaluating the expression:
d ≈ 2.4507 ft
Therefore, the numerical value of the distance you slide the puck up the ramp is approximately 2.4507 feet.