Position (m) Force (N) Acceleration (m/s 2 ) 0.100 0.050 0 ? 0.050 ? 0.100 Use t

Question

Position (m) Force (N) Acceleration (m/s2) 0.100 0.050 0 ?0.050 ?0.100 Use the values from above For the same spring and mass system, find the force exerted by the spring and the position x when the object's acceleration is +8.65 m/s2. F = N x = cm Use the values from above For the same spring and mass system, find the force exerted by the spring and the position x when the object's acceleration is +8.65 m/s2. F = N x = cm Position (m) Force (N) Acceleration (m/s2) 0.100 0.050 0 ?0.050 ?0.100 A 0.310 kg object attached to a spring of force constant 1.40 times 02 N/m is free to move on a frictionless horizontal surface. If the object is released from rest at x = 0.100 m, find the force on it and its acceleration at x = 0.100 m, x = 0.0500 m, x = 0 m,x = ?0.0500 m, and x = ?0.100 m. For the same spring and mass system, find the force exerted by the spring and the position x when the object's acceleration is +8.65 m/s2.

Explanation / Answer

AT POSITION(X)

FORCE = (-K*X)

ACCELERATION = (FORCE/M)

X = O.100

FORCE = -1.4 * 10^2 *0.100 = -1.4*10^4

ACCELERATION = -1.4*10^4 / 0.310 = = - 4.5*10^4

X = 0.05

FORCE = -1.4 * 10^2 *0.050 = -7*10^4

ACCELERATION = -7*10^4 / 0.310 = - 22.55*10^4

X = 0

FORCE= 1.4 * 10^2 * 0 = 0

ACCELERATION = 0 / 0.310 = 0

X = -0.05

FORCE = -1.4 * 10^2 *-0.050 = 7*10^4

ACCELERATION = 7*10^4 / 0.310 = 22.55*10^4

X = -O.100

FORCE = -1.4 * 10^2 *-0.100 = 1.4*10^4

ACCELERATION = 1.4*10^4 / 0.310 = 4.5*10^4

FORCE = M*ACCELERATION

= 0.310*8.65 = 2.6816

FORCE = -K*X

X = -FORCE/K = - 2.6816/(1.4 * 10^2) = - 0.019 m

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