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Two identical massless springs of constant k = 200 N/m are fixed at opposite end

ID: 1980823 • Letter: T

Question

Two identical massless springs of constant k = 200 N/m are fixed at opposite ends of a level track, as shown below. A 5.00 kg block is pressed against the left spring, compressing it by 0.150 m. The block (initially at rest) is then released, as shown in Figure (a). The entire track is frictionless except for the section between A and B.

(a) Determine the maximum compression of the spring on the right [see (b)].
(b) Determine where the block eventually comes to rest, as measured from A [see (c)].

Two identical massless springs of constant k = 200 N/m are fixed at opposite ends of a level track, as shown below. A 5.00 kg block is pressed against the left spring, compressing it by 0.150 m. The block (initially at rest) is then released, as shown in Figure (a). The entire track is frictionless except for the section between A and B. Given that the coefficient of kinetic friction between block and track along AB is mu k = 0.0800,and given that the length AB is 0.250 m, (a) Determine the maximum compression of the spring on the right [see (b)]. (b) Determine where the block eventually comes to rest, as measured from A [see (c)].

Explanation / Answer

The energy imparted to the block by the left spring is 0.5*k*?x² The energy absorbed by the frictional patch is F*d. F = µ*m•g. The energy of the block when it reaches the right side spring is then 0.5*k*?x² - µ*m•g all of this energy is transferred to the right side spring at its full compression ?y, so 0.5*k*?y² = 0.5*k*?x² - µ*m•g ?y² = ?x² - 2*µ*m•g/k The block will come to rest when the total distance traveled across the frictional part takes all the energy imparted by the left spring (nothing else in this system removes energy from the block). Thus Fr*s = 0.5*k*?x² µ*m*g*s = 0.5*k*?x² s = 0.5*k*?x² / µ*m*g then divide s by the length of the path to determine how many times the block travels the length of the frictional area. If A is the entry point of the block to that area, you can then determine where the block stops. Using the numbers k = 200 N/m, ?x = 0.18 m, µ=0.0800 and m = 3.5 kg, I get s = 1.181 m That divided by 0.265 m is 4.457 m The block traverses the patch 4 times ending at A. It then advances 0.457*0.265 m = 0.121 m past A

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