Ever since you were young, you’ve been famous for playing a mean game of pinball

Question

Ever since you were young, you’ve been famous for playing a mean game of pinball. You draw back the launcher spring a distance of x = 13cm and release to shoot the pinball horizontally along the table. The spring constant is k = 34 N/m. Assume the pinball slides without rolling (no friction) and has a mass of m = 80g.

a. How many pounds of force is required to pull the launcher back?

b. What is the velocity of the ball at the moment it leaves the launcher?

c. After release, the ball rolls up a 30° ramp. What is the highest height h it could attain? Answer first in terms of variables, then find the number.

d. If instead of being frictionless, the ramp had a rough surface with a coefficient of friction of = 0.25, how would your answer to part c change?

Explanation / Answer

part a:

force required=spring constant*compressed distance of the spring

=34*0.13

=4.42 N

converting from newton to pound force, we get force required=0.99365 pounds.

part b:

the entire potential energy of the spring will be converted to kinetic energy of the ball.

if speed of the ball at the moment it leaves the launcher is v m/s,

then 0.5*m*v^2=0.5*k*x^2

==>v=sqrt(k*x^2/m)=2.68 m/s

part c:

when it attains highest height, the initial kinetic energy of the ball will be completely converted to potential energy

as there is no energy loss due to friction.

hence 0.5*m*v^2=m*g*h

==>h=0.366 m

part d:

let height attained is h meters.

then distance travelled along the ramp is h/sin(30)=2*h

friction force=friction coefficient*normal force

=0.25*m*g*cos(30)

=0.16974 N

then work done against friction=friction force*distance travelled=0.16974*2*h joules

using energy conservation principle,

initial kinetic energy-work done against friction=final potential energy

==>0.5*m*v^2-0.16974*2*h=m*g*h

==>h=0.2557 m

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