1/2 m v^2 = 1/2 k x^2
14 * 2.9^2 / 4884 = x^2
x = 0.1553 ... 0.16 to two sig fig ... might be a rounding error
A mass m = 14 kg rests on a frictionless table and accelerated by a spring with spring constant k = 4884 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.52. The mass leaves the spring at a speed v = 2.9 m/s.
1) How far was the spring stretched from its unstreched length?
I tried this but it still wrong...
58.8=(0.5)(4884)x^(2)
x=0.15 , -0.15
Can you help me with this?
1 answer
1 answer