Recent content by Symon

  1. S

    Work on a baseball being caught by a catcher

    I would say the stopping force acts in the opposite direction that the ball is thrown, whilst the glove moves forewards in the same direction as the ball. Correct? Or have i gotten myself all backwards haha.
  2. S

    Work on a baseball being caught by a catcher

    Does it imply a change in the displacement vector?
  3. S

    Work on a baseball being caught by a catcher

    Ah I understand, so when they ask for me to find "the average stopping force" are they asking me to re-arrange the W=F x d formula to solve for the force? I'm really sorry about all the stupid questions I'm just trying to wrap my head around everything.
  4. S

    Work on a baseball being caught by a catcher

    Is the theorem Work = Ekfinal - Ekinitial? If the initial kinetic energy = 5J, the final should be 0Jm because you're multiplying by 0 m/s. This gives me -5J which throws me off because I don't know if I should be taking the absolute value or not.
  5. S

    Work on a baseball being caught by a catcher

    Homework Statement A baseball of mass 250g (0.25kg), pitched with a speed of 40m/s, is caught by a catcher whose glove moves backward 0.25m while stopping the ball. a) What is the kinetic energy of the ball? (solved) b) How much work did the catcher's glove do on the ball? c) What was the...
  6. S

    Finding the displacement of a braking car, with limited variables.

    Thus far, i am unfamiliar with that equation, here are some of the equations that I've either derived or have been given: 1) ∆d=〖(∆v)〗^2/2a 2) ∆d= ∆v∆t 3) ∆d=vi∆t+ 1/2 a〖(∆t)〗^2 As you can see, the first equation that i have neglects time, but i am unable to use it without finding the...
  7. S

    Finding the displacement of a braking car, with limited variables.

    Homework Statement A car is traveling at 120km/h when it slams on the brakes. How long is the skid mark if the coefficient of friction is 0.62?(hint: convert km/h to m/s) Given: initial velocity: 33.3m/s; final velocity: 0m/s; coefficient of fricition: 0.62 Homework Equations μ =...
  8. S

    What is the tension on the elevator cable?

    a) For when the elevator is accelerating upwards at 1.2m/s^2: Tension = m(g+a) = 2.7x10^3kg (9.8m/s^2 [upwards] + 1.2m/s^2 [upwards]) Is this correct? I'm trying to interpret this; g = 9.8m/s^2 [UP] because the Tension is acting opposite to gravities force on the elevator? This would...
Back
Top