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A block of mass 0.18 kg is attached to a spring with force constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially at rest with the spring unstretched. An impulse sets the block moving. The block slides 0.06 m and then momentarily stops for the first time. The initial speed of the block is V = N/10 m/s. Find N. (Take g = 10 m/s²)
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- 2
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- 4
Correct answer: 4
Solution
Energy conservation from start to when block stops: KE_initial = PE_spring + W_friction. (1/2)(0.18)v² = (1/2)(2)(0.06)² + (0.1)(0.18)(10)(0.06). 0.09*v² = (1)(0.0036) + 0.0108 = 0.0144. v² = 0.0144/0.09 = 0.16. v = 0.4 m/s = 4/10 m/s. So N = 4. Wait: let me redo. (1/2)(0.18)v² = (1/2)(2)(0.0036) + (0.018)(0.06). = 0.0036 + 0.00108... Recompute: (1/2)*k*x² = 0.5*2*(0.06)² = 1*(0.0036) = 0.0036 J. Friction work = mu*m*g*x = 0.1*0.18*10*0.06 = 0.0108 J. KE = 0.0036+0.0108=0.0144 J. (1/2)(0.18)v²=0.0144. 0.09*v²=0.0144. v²=0.16. v=0.4 m/s. N=4. But options show 1,2,3,4 and answer N=4: V=4/10=0.4 m/s.
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