A 0.35 kg object attached to the spring force constant 1.3 x 102 N/m is free to

Question

A 0.35 kg object attached to the spring force constant 1.3 x 102 N/m is free to move on a frictionless horizontal surface. If the object is released from rest at x = 10 cm, find its acceleration.
A) -30 m/s2 B) 3714 m/s2 C) -130 m/s2 D) 50 m/s2 E) -37 m/s2 A 0.35 kg object attached to the spring force constant 1.3 x 102 N/m is free to move on a frictionless horizontal surface. If the object is released from rest at x = 10 cm, find its acceleration.
A) -30 m/s2 B) 3714 m/s2 C) -130 m/s2 D) 50 m/s2 E) -37 m/s2
A) -30 m/s2 B) 3714 m/s2 C) -130 m/s2 D) 50 m/s2 E) -37 m/s2

Explanation / Answer

Acceleration is given by the formula

|a| = w^2*x ( magnitude)

Here , w is angular velocity and x is displacement

Since , w = (k/m)

w^2 = k/m

so, a = k*x/m

given , m = 0.35 kg , k = 1.3*10^2 N/m , x = 0.1 m

a = 1.3*10^2*0.1/0.35 = 37.14 m/s^2

a = 37.14 m/s^2 (equivalent to 37 m/s^2)

answer) option E ) 37 m/s^2

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