A light spring of force constant 4.25N/m is compressed by 8.00 cm and held betwe

Question

A light spring of force constant 4.25N/m is compressed by 8.00 cm and held between a 0.250 kg block onthe left and a 0.420 kg block on theright. Both blocks are at rest on a horizontal surface. The blocksare released simultaneously so that the spring tends to push themapart. Find the maximum velocity each block attains if thecoefficient of kinetic friction between each block and the surfaceis the following. In each case, assume that the coefficient ofstatic friction is greater than the coefficient of kineticfriction. Let the positive direction point to the right. µk 0.250 kg block 0.420 kg block 0.000 1 m/s 2 m/s 0.100 3 m/s 4 m/s 0.491 5 m/s 6 m/s
µk 0.250 kg block 0.420 kg block 0.000 1 m/s 2 m/s 0.100 3 m/s 4 m/s 0.491 5 m/s 6 m/s

Explanation / Answer

Given :    The spring constant, or k, of the spring is given as 4.25N/m. M1 (the block on the left)=0.250kg M2 (the block on the right)=0.420kg x = 8 cm = 8 * 10-2 m You are asked to find the acceleration as the blocks as theyare released, with variable friction on the horizontal surface theyare on. WHEN THE COEFFICIENT OF KINETICFRICTION=0 By drawing a free body diagram of each block, you find thatwithout the influence of friction, the net force on each block issimply the force provided by the spring, or F=kx.Substituting this into the equation F=ma, we find that: RIGHT BLOCK:    k x = m a ( 4.25 N/m) (8.00 * 10-2 m)=(0.420 kg) (a) 0.308 = 0.5a a=0.8 m/s2 LEFT BLOCK: kx=ma ( 4.25 N/m) (8.00 * 10-2 m)=(0.250 kg) (a) a= - 1.36 m/s2 Please note that this acceleration is negative due to itsdirection (left) WHEN THE COEFFICIENT OF KINETICFRICTION=0.100 By drawing the free-body diagram for each block, you find thatwith the added influence of friction, the net force on each blockis F = kx - kmg. Substituting this into themain equation F=ma, we find that: RIGHT BLOCK:    (k x-mg) = ma ((4.25 N/m)(8..0 * 10-2m) - (0.1) (0.420kg)(9.8)) =(0.420kg)(a) a = ------ m/s2 We can deduce, because the acceleration that we would recievefrom this is negative, that the box would not move at all. Thus,the acceleration can simply be thought of as0. LEFT BLOCK:    (k x-mg) = ma ((4.25 N/m)(8..0 * 10-2m) - (0.1) (0.250kg)(9.8)) =(0.250kg)(a) a = ------ m/s2     The lighter block, unlike the heavier one,does begin to move. Please note, however, that the acceleration isnegative due to its direction (left). WHEN THE COEFFICIENT OF KINETICFRICTION=0.491 Like the last friction value, the net force on each block isF = kx - kmg. Substituting this into themain equation F=ma, we find that: RIGHT BLOCK:    (k x-mg) = ma ((4.25 N/m)(8..0 * 10-2m) - (0.491) (0.420kg)(9.8)) =(0.420kg)(a) a = ------ m/s2 We can deduce, because the acceleration that we would recievefrom this is negative, that the box would not move at all. Thus,the acceleration can simply be thought of as0. LEFT BLOCK:   (k x-mg) = ma ((4.25 N/m)(8..0 * 10-2m) - (0.1) (0.250kg)(9.8)) =(0.250kg)(a) a = ------ m/s2 We can deduce, because the acceleration that we would recievefrom this is negative, that the box would not move at all. Thus,the acceleration can simply be thought of as 0. Solve it I hope it helps you The spring constant, or k, of the spring is given as 4.25N/m. M1 (the block on the left)=0.250kg M2 (the block on the right)=0.420kg x = 8 cm = 8 * 10-2 m You are asked to find the acceleration as the blocks as theyare released, with variable friction on the horizontal surface theyare on. WHEN THE COEFFICIENT OF KINETICFRICTION=0 By drawing a free body diagram of each block, you find thatwithout the influence of friction, the net force on each block issimply the force provided by the spring, or F=kx.Substituting this into the equation F=ma, we find that: RIGHT BLOCK:    k x = m a ( 4.25 N/m) (8.00 * 10-2 m)=(0.420 kg) (a) 0.308 = 0.5a a=0.8 m/s2 LEFT BLOCK: kx=ma ( 4.25 N/m) (8.00 * 10-2 m)=(0.250 kg) (a) a= - 1.36 m/s2 Please note that this acceleration is negative due to itsdirection (left) WHEN THE COEFFICIENT OF KINETICFRICTION=0.100 By drawing the free-body diagram for each block, you find thatwith the added influence of friction, the net force on each blockis F = kx - kmg. Substituting this into themain equation F=ma, we find that: RIGHT BLOCK:    (k x-mg) = ma ((4.25 N/m)(8..0 * 10-2m) - (0.1) (0.420kg)(9.8)) =(0.420kg)(a) a = ------ m/s2 We can deduce, because the acceleration that we would recievefrom this is negative, that the box would not move at all. Thus,the acceleration can simply be thought of as0. a = ------ m/s2 We can deduce, because the acceleration that we would recievefrom this is negative, that the box would not move at all. Thus,the acceleration can simply be thought of as0. LEFT BLOCK:    (k x-mg) = ma ((4.25 N/m)(8..0 * 10-2m) - (0.1) (0.250kg)(9.8)) =(0.250kg)(a) a = ------ m/s2 ((4.25 N/m)(8..0 * 10-2m) - (0.1) (0.250kg)(9.8)) =(0.250kg)(a) a = ------ m/s2 a = ------ m/s2     The lighter block, unlike the heavier one,does begin to move. Please note, however, that the acceleration isnegative due to its direction (left). WHEN THE COEFFICIENT OF KINETICFRICTION=0.491 Like the last friction value, the net force on each block isF = kx - kmg. Substituting this into themain equation F=ma, we find that: RIGHT BLOCK:    (k x-mg) = ma ((4.25 N/m)(8..0 * 10-2m) - (0.491) (0.420kg)(9.8)) =(0.420kg)(a) a = ------ m/s2 ((4.25 N/m)(8..0 * 10-2m) - (0.491) (0.420kg)(9.8)) =(0.420kg)(a) a = ------ m/s2 a = ------ m/s2 We can deduce, because the acceleration that we would recievefrom this is negative, that the box would not move at all. Thus,the acceleration can simply be thought of as0. LEFT BLOCK:   (k x-mg) = ma ((4.25 N/m)(8..0 * 10-2m) - (0.1) (0.250kg)(9.8)) =(0.250kg)(a) a = ------ m/s2 ((4.25 N/m)(8..0 * 10-2m) - (0.1) (0.250kg)(9.8)) =(0.250kg)(a) a = ------ m/s2 a = ------ m/s2 We can deduce, because the acceleration that we would recievefrom this is negative, that the box would not move at all. Thus,the acceleration can simply be thought of as 0. Solve it I hope it helps you Solve it I hope it helps you

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