A 7.25 kg block starts at the top of a 13.6 m long incline that has an angle of
ID: 1475957 • Letter: A
Question
A 7.25 kg block starts at the top of a 13.6 m long incline that has an angle of 10 to the horizontal. The block then slides out on a horizontal frictionless surface and collides with a 7.11 kg block in an inelastic collision in which the blocks stick together. The blocks then slide to the right onto a frictional section of track as a result of the collision.
What was the velocity of the 7.25kg block at the bottom of the ramp?
How much Kinetic Energy was lost in the collision?
How far do the blocks slide to the right on the frictional surface before stopping if the coefficient of kinetic friction is k = 0.18.
Explanation / Answer
mgLsin(10) = 0.5*mV^2
2*gLsin(10)=V^2
2*9.81*13.6*sin(10)
V= 6.80697366994 m/s
after collision
speed = 7.25*6.80697366994 /(7.25+7.11) =3.43666846149 m/s
KE lost = 0.5* (7.25+7.11)(7.25*6.80697366994 /(7.25+7.11))^2 - 0.5*7.25 *6.80697366994 ^2
=83.1632231991 Joules
qriting NLM 1
0= 3.43666846149 - 0.18g t
t=1.94623879346s
S = 0.5*0.18g t^2
=0.5*0.18*9.81*1.94623879346^2
=3.34428874001 m
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