Recent content by murrskeez

  1. M

    Euler's method for a mass sliding down a frictionless curve

    V2=1.96+9.8(0.2)=3.92 V3=3.92+9.8*cos(X3/2)=5.842473 I can use excel to finish it off if I have the right pattern.
  2. M

    Euler's method for a mass sliding down a frictionless curve

    V0=0 V1=0+9.8*0.2=1.96 X0=0 X1=0+0(0.2)=0 X2=0+1.96(0.2)=.392
  3. M

    Euler's method for a mass sliding down a frictionless curve

    Ok so for the sum of the radial components of the applied forces I am getting: ΣFr=m(v2/R) N-mgsin(θ)=m(v2/R) I'm not sure where to go from here though.
  4. M

    Euler's method for a mass sliding down a frictionless curve

    Homework Statement Consider a mass sliding down a frictionless curve in the shape of a quarter circle of radius 2.00 m as in the diagram. Assume it starts from rest. Use Euler’s method to approximate both the time it takes to reach the bottom of the curve and its speed at the bottom. Hint...
  5. M

    The Work Done on a Sliding Box: How to Calculate Friction and Applied Force

    Homework Statement A box of mass m is sliding along a horizontal surface. Part A) The box leaves position x=0 with a speed v0. The box is slowed by a constant frictional force until it comes to rest at a position x=x1. Find Ff, the magnitude of the average frictional force that acts on the...
  6. M

    Roller Coaster Physics in a Gravity-Less Park?

    Homework Statement A roller coaster is designed in an area of the park that is gravity-less. The roller coaster will accelerate a cart with a total mass of 10kg, from the initial position as shown in the figure to point A. After passing through point A, the cart will no longer speed up and...
  7. M

    Standing Waves on a string & pipe

    I think I get it :smile: fpipe = nV/2L 706.5Hz = (3*344m/s)/(2L) L = 0.730m fo = v/2L fo = (344m/s)/(2*0.730m) fo = 236Hz thank you so much, really appreciate it :)
  8. M

    Standing Waves on a string & pipe

    So the third overtone would mean n=4... so fn=(4*141.3m/s)/(2*0.400m) = 706.5Hz so the frequencies of the string and pipe must be related, I'm just not sure how.
  9. M

    Standing Waves on a string & pipe

    Homework Statement A string 40.0cm long of mass 8.50g is fixed at both ends and is under a tension of 425N. When the string is vibrating in its third overtone, you observe that it causes a nearby pipe, open at both ends, to resonate in its third harmonic. The speed of sound is 344m/s. a) How...
  10. M

    Using Energy to Solve Work Problem

    Finally got the correct answer, thank you very very much :smile:
  11. M

    Using Energy to Solve Work Problem

    Oops, looks like I dropped the s. Looks like that would make y = sμk(gcos15°)(-1) + 1/2(11.0)2 / 9.8 m/s2 could I then relate s and y through sin? Am I even on the right track? The first equation is Wother = ΔE , other in this case meaning friction. Was given to me by my professor. But...
  12. M

    Using Energy to Solve Work Problem

    I thought that the in the formula W = sfcos∅ , the angle ∅ should represent the angle between the direction of s and the corresponding force, which in this case would be friction...and those two directions are 180° apart, right?
  13. M

    Using Energy to Solve Work Problem

    Homework Statement A truck is traveling at 11.1 m/s down a hill when the brakes on all four wheels lock. The hill makes an angle of 15.0° with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is 0.750. How far does the truck travel before coming to...
Back
Top