The spring in the figure has a spring constant of 1400N/m . It is compressed 15.

Question

The spring in the figure has a spring constant of 1400N/m . It is compressed 15.0cm , then launches a 200g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.180.

Q1:What distance d does the block sail through the air?

I followed every steps here http://www.chegg.com/homework-help/questions-and-answers/spring-figure-spring-constant-1400-n-m-compressed-150-cm-launches-200-g-block-horizontal-s-q2356145

I got 12.73m which is incorrect. what's the right answer?

The spring in the figure has a spring constant of 1400N/m . It is compressed 15.0cm , then launches a 200g block. The horizontal surface is frictionless, but the block's coefficient of kinetic friction on the incline is 0.180. Q1:What distance d does the block sail through the air?

Explanation / Answer

k = spring constant = 1400 N/m
x = 15 cm =0.15 m
m=mass = 200g = 0.2 kg
u =coefficient of kinetic friction on the incline is 0.180.
PE = potential energy of the spring = 0.5kx^2
KE = kinetic energy of the block at the bottom of the incline = 0.5mv^2
PE=KE
0.5kx^2 = 0.5mv^2
v= 10.607 m/s
z= angle of the slope = 45
sum forces perpendicular to plane
N - mg cos z =0
N= mg cos z
sum forces parallel to plane
F- f = ma
- uN = ma
-u mg cos z = ma
a = -1.526 m/s^2
Vf= final velocity
v = initial velocity=10.607 m/s
s= length of the slope = s= a^2 + b^2 a=b so
s=?2(2^2)=2.83

Vf^2 = v^2 +2as
Vf =?(v^2 +2as) = 10.19 m/s

use the range equation R = Vf^2/g sin (2z)
R= 10.59 m

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