A 5 kg box is attached to a spring that has an elastic coefficient of 110 N/m. How much force is required to stretch or compress the spring one meter?

Ibelieve the answer is either b or c, but i'm leaning more towards c.hope my answer has you : )

Special care would be an answer

Friction force is equal to the net force of the car.             fn                           |ff< ·                       |                     fgfg and fn cancel out.55 km/h is 15.2778 m/s. this is the initial velocityfinal velocity is 0 m/sfind acceleration using kinematic equationv² = v₀² + 2adsince v = 00 = v₀² + 2ada = -v₀²/(2d)a = -(15.2778 m/s)/(2 * 38m) a = -3.2418 m/s²find net force and you find friction.f = ma f = (1165 .2418 m/s²)f = 3776.7 n = ffw = ff · dw = (3776.7 n)(38 m)w = -1.4 × 10^5 jthe friction forces do 1.4 × 10^5 j of work on the car