A light spring of force constant 3.85 N/m is compressed by 8 cm and held between

Question

A light spring of force constant 3.85 N/m is compressed by 8 cm and held between a .250 kg block on the left and a .5 kg block on the right. Both blocks are at rest on a horizontal surface. The blacks are released simultaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is a)0, b).1, and c).462. Assume that the coefficient of static friction is larger than that for kinetic friction.

Explanation / Answer

Given: force constent of the spring K = 3.85 N/m distance x = 8*10-2 m mass of the block m1 = 0.25 kg mass of the block m2 = 0.5 kg .......................................................................................................... (a) In first case friction is zero now from the conservation of energy 1/2 kx2 = 1/2mv2 for block 1 1/2 kx2 = 1/2mv12 v1 = sqrt(k/m)x      = sqrt [(3.85 N/m) / 0.25 kg] 8*10-2 m      = 0.31 m/sec for block 2 1/2 kx2 = 1/2mv12 v1 = sqrt(k/m)x      = sqrt [(3.85 N/m) / 5 kg] 8*10-2 m      = 0.07 m/sec ....................................................................... (2) now the coeffiecient of friction = 0.1 now from conservation of energy 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.1)( 0.25 kg)(9.8m/sec2) v1 = 0.616 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.1)( 0.5 kg)(9.8m/sec2) v2 = 1.38 m/sec (3) now the coeffiecient of friction = 0.4 now from conservation of energy 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.4)( 0.25 kg)(9.8m/sec2) v1 = 2.78 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.4)( 0.5 kg)(9.8m/sec2) v2 = 2.79 m/sec for block 2 1/2 kx2 = 1/2mv12 v1 = sqrt(k/m)x      = sqrt [(3.85 N/m) / 5 kg] 8*10-2 m      = 0.07 m/sec ....................................................................... (2) now the coeffiecient of friction = 0.1 now from conservation of energy 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.1)( 0.25 kg)(9.8m/sec2) v1 = 0.616 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.1)( 0.5 kg)(9.8m/sec2) v2 = 1.38 m/sec (3) now the coeffiecient of friction = 0.4 now from conservation of energy 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.4)( 0.25 kg)(9.8m/sec2) v1 = 2.78 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.4)( 0.5 kg)(9.8m/sec2) v2 = 2.79 m/sec ....................................................................... (2) now the coeffiecient of friction = 0.1 now from conservation of energy 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.1)( 0.25 kg)(9.8m/sec2) v1 = 0.616 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.1)( 0.5 kg)(9.8m/sec2) v2 = 1.38 m/sec (3) now the coeffiecient of friction = 0.4 now from conservation of energy 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.4)( 0.25 kg)(9.8m/sec2) v1 = 2.78 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.4)( 0.5 kg)(9.8m/sec2) v2 = 2.79 m/sec 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.1)( 0.25 kg)(9.8m/sec2) v1 = 0.616 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.1)( 0.5 kg)(9.8m/sec2) v2 = 1.38 m/sec 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.1)( 0.5 kg)(9.8m/sec2) v2 = 1.38 m/sec (3) now the coeffiecient of friction = 0.4 now from conservation of energy 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.4)( 0.25 kg)(9.8m/sec2) v1 = 2.78 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.4)( 0.5 kg)(9.8m/sec2) v2 = 2.79 m/sec now the coeffiecient of friction = 0.4 now from conservation of energy 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.4)( 0.25 kg)(9.8m/sec2) v1 = 2.78 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.4)( 0.5 kg)(9.8m/sec2) v2 = 2.79 m/sec 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.25 kg)(v12 ) + (0.4)( 0.25 kg)(9.8m/sec2) v1 = 2.78 m/sec for second mass m2 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.4)( 0.5 kg)(9.8m/sec2) v2 = 2.79 m/sec 1/2 kx2 = 1/2mv12 + mg ....... (1) 1/2*(3.85 N/m)( 8*10-2 m)2 = 1/2( 0.5 kg)(v12 ) + (0.4)( 0.5 kg)(9.8m/sec2) v2 = 2.79 m/sec

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