In the diagram above, an object of mass m = 2.50kg is sitting on a frictionless
ID: 1454846 • Letter: I
Question
In the diagram above, an object of mass m = 2.50kg is sitting on a frictionless inclined plane which makes an angle of 60 degree with the horizontal (ground). The distance from m to the ground (along the plane) is 11.547 meters. The distance from point g to the spring (gs) is 50 meters. The object is allowed to slide down the plane, with no initial velocity, (V_0 = 0), and we would like to follow its motion from the time(t_0) it starts to slide. a) What is the potential energy of the object at this point in time (t_0)? b) What is the velocity when it reaches the ground? c) Assuming no friction on the ground, with what velocity does it strike the spring? d) By what distance is the spring compressed? (e,f) Repeat parts (c) and (d) if the coefficient of friction between the object and the ground is 0.1? g) What is the maximum distance of the spring from point g, so that the object actually makes it all the way to the spring? h) Repeat part (c) if the initial velocity is 20 m/s, (Once again assuming that the friction is zero everywhere.) i) How does the answer to (h) change if the initial velocity is up the plane instead of down the plane?Explanation / Answer
(a) height(h) at t=t0 ; h=11.547*sin(60) =10
Potential energy= mgh = 245 J
b) Apply conservation of energy
mgh= mv2/2 ; v=14 ms-1
c) it strikes with v=14 ms-1 as before
d) when the spring compresses maximumly, potentail energy = kx2/2 and kinetic energy=0 at that point
Apply Conservation of energy
mv2/2 = kx2/2 ;
distance compressed by(x)= 0.099 m