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brown20b
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Homework Statement
A 5.0 kg mass is released from rest, at the top of a frictionless 30° inclined plane. Calculate the kinetic energy of the mass when it has slid 40cm down the incline starting from rest. (ANS = 9.80 J)
Homework Equations
KE=1/2(mv^2)
a=v/t
d=v1t + 1/2a(t^2)
v2^2=v1^2 + 2ad
v2=v1 + at
d= (v1+v2/2)*t
The Attempt at a Solution
G: v1= 0 ; θ= 30° ; d=40cm
R: KE=? ; v2=? ; t=?
(NE direction is +ve)
(1) v2y^2=v1y^2 - 2aydy
v2y^2= √0 -2(9.80)(-0.40sin30)
v2y= 1.98 m/s
(2) v2y=v1y +at
t = v2y/ay
t = 0.202 s
(3) v2x= dx/t
v2x= 0.40cos30/0.202
v2x= 1.71 m/s
(4) v2= √(1.98)^2 + (1.71)^2
v2= 2.62 m/s
(5) KE = 1/2(mv^2)
KE = 1/2(5.0*2.62^2)
KE = 17.2 J
My answer is far off from the correct one so I'm assuming my approach might be wrong.. But I just don't know from where it is that I'm doing it wrong.. Probably something with my calculations, not too sure entirely..