A small glider is placed against a compressed spring at the bottom of an air tra

Question

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 30.0 degree above the horizontal. The glider has mass 9.00 times 10-2kg. The spring has 630N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.90m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring. Calculate the gravitational potential energy of the system when the glider is at the maxium distance up the slope. What distance was the spring originally compressed? When the glider has traveled along the air track 0.900m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Explanation / Answer

.theta = 30o
M = 6900 * 10^-2 kg
k = 630 N/m
Vertical height travelled h = 1.90 m * sin(theta) = 1.90 m * sin( 30o) = 0.95 m
By conservation of energy,
1/2 * k * x^2 = Mgh
1/2 * 630 * x^2 = 9.00 * 10^-2 * 9.8 * 0.95
315 * x^2 = 0.8379
x^2 = 0.8379/315 = 26.6 * 10^-4
x = sqrt(26.6 * 10^-4) = 5.1575 * 10^-2 m
x= 5.158 * 10^-2 m

a) gravitational Potential Energy at the max distance ,PE= m g h = 9.0*10-2*9.80*1.9*sin(30o)= 0.839 J

b)As calculated above for distance for which the spring was compressed, x =5.158 * 10-2 m

c) Now since that more than the compression amount

d) Initial KE + initial PE (gravitational)= final KE + final PE (gravitational)

0+ 0.839 = final KE + 9.0*10-2*9.80*0.9*sin(30o)

final KE = 0.839 - 9.0*10-2*9.80*1.9*sin(30o)=0.442 J

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