A box of mass 1.00 kgstarts from rest at the top (point A) of a frictionless inc
ID: 1753122 • Letter: A
Question
A box of mass 1.00 kgstarts from rest at the top (point A) of a frictionless inclineof
height 2.00 m (see figurebelow). (a) Use energy methods to find the speed of thebox
when it reaches the bottomof the ramp (point B). (b) Use Newton’s second law tofind
the acceleration of thebox, then use it to determine the speed of the box at the bottomof
the ramp ifq =30°.(c) What is the speed of the box at the point labeled C? Explain.(d)
When the box reaches thefar end of the track it encounters a rough patch withk= 0.500.
Use energy methods todetermine the distance that the box travels along the roughpatch
before it comes to rest(point D). (e) Repeat part d, but now use Newton’slaws.
Explanation / Answer
NOTE: in my solution im using g = 9.8 m/s^2, depending on yourbook, you may be used to using g = 10 m/s^2. to answer part A, we see total potential energy converts tokinetic energy... mgh = 1/2mv^2 mass cancels out gh = 1/2v^2 v = square root (2gh) v = square root (2*9.8*2m) v = 6.26m/s applying newton's second laws... mgsin = ma gsin = a a = 9.8sin30 a = 4.9 m/s^2 Using this to find speed... V^2 = Vo^2 + 2ax we can determine x(the hypotenuse of theincline) by looking at simple triginmetric relations... sin = opp / hyp sin = h/x h/sin(30) =x = 4m going back to our equation... V^2 = 0 + 2(4.9)(4m) V = 6.62 m/s (notice our same result in step 1, theta isindependent of speed, solely depends on initial potentialenergy) At point C, gravity is still the only force(gravity is aconservative force independent of path), it is still frictionless,so mgh = 1/2mv^2, same expression takes place , and the speed isstill 6.62 m/s. W = k W = -1/2(1kg)(6.62)^2 W = -21.91 J W = F*d Where Friction is defined as K * N Where normal force... N-mg = 0 N = mg N = 9.8 N so.. Friction = K * N Friction = .500 * 9.8 = 4.9 N = -4.9 N ( opposing motion so weput negative sign) W = F*d -21.91J = -4.9 d d = 4.47 m lets use newton's laws to validate this as the problemasks... -Fric = ma -4.9N = (1kg) a -4.9 m/s^2 = a using kinematics... V^2 = V0^2 + 2ax 0 = (6.62^2) + 2(-4.9) x x= 4.47 m The reason Vo is still 6.62 m/s is because we agreed thatbefore the friction came, the velocity would not change..due to theconservation of energy ( ideally ) :P Any questions, just ask Please rate me well , this took some time to type =pRelated Questions
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