The spring in the figure has a spring constant of 1000 N / m . It is compressed
ID: 2252473 • Letter: T
Question
The spring in the figure has a spring constant of 1000N/m . It is compressed 14.0cm , then launches a 200g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.210.
What distance d does the block sail through the air?
Explanation / Answer
firstly calculate the velocity
v^2 = (k.x^2) / m = (1000 x 0.14^2)/0.2 = 98
v= 9.89 m/s
then equation of motion is
-m.g.sin(p) - Um.g.cos(p) = ma
a = -g.sin(p) - U.g.cos(p)
a= -9.8* sin( 45)- 0.2 * 9.8* cos(45)
a= -9.36 m/s^2
Distance up plane = 2/sin(45) = 2.83 m
Speed at top of plane
v^2 = u^2 + 2as = 9.89^2 - 2*9.36* 2.83
v^2 = 44.83
v = 6.69 m/s
t = (6.69x sin(45))/9.81 = 0.58 s
Horizontal distance travelled in this time
6.69 cos(45) x 0.58 = 6.05 m
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