A 0.374-kg block is attached to ahorizontal spring that is at its equilibrium le

Question

A 0.374-kg block is attached to ahorizontal spring that is at its equilibrium length, and whoseforce constant is 16.8 N/m. The blockrests on a frictionless surface. An arrow with a mass of 0.0500 kgis shot horizontally at the block, hitting it with an initial speedof 1.98 m/s and sticking. How far does theblock-arrow system compress the spring?
A 0.374-kg block is attached to ahorizontal spring that is at its equilibrium length, and whoseforce constant is 16.8 N/m. The blockrests on a frictionless surface. An arrow with a mass of 0.0500 kgis shot horizontally at the block, hitting it with an initial speedof 1.98 m/s and sticking. How far does theblock-arrow system compress the spring?

Explanation / Answer

Given M = 0.374 kg            k =16.8 N/m            m =0.050 kg            u =1.98 m/s The formula is m*u = ( M + m ) v                             v = m*u / ( M + m) =0.05*1.98/(0.374+0.05)=0.233                               From the conservation of energy                     (1/2)(M+m) v2 = (1/2) k x2                                         x = v * [ (M+m) / k ]1/2          =0.233*sqrt((0.374+0.05)/16.8)=0.037

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