(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

This problem has actually been on here twice (with varying numbers) but not yet solved entirely

A 30.0-kg block is resting on a flat horizontal table. On top of this block is resting a 15.0-kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is 345 N/m. The coefficient of kinetic friction between the lower block and the table is 0.640, while the coefficient of static friction between the two blocks is 0.940. A horizontal force F is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed.

At the point where the upper block begins to slip on the lower block determine the following.

(a) the amount by which the spring is compressed

(b) the magnitude of the force F

2. Relevant equations

[tex] F = kx [/tex]

[tex]f = \mu F_N [/tex]

[tex]F = ma [/tex]

3. The attempt at a solution

a) correctly solved by me (with no assistance)

[tex] F_{spring} \geq f_{UL} [/tex]

[tex]k x \geq \mu_{UL} m_u g [/tex]

[tex]x \geq \frac{\mu_{UL} m_u g}{k} , k \gt 0 [/tex]

plugging in my values, 0.40m

b) most logically I need to figure out how [tex] F \propto f_{UL} [/tex] .. I can't quite figure out the interaction between the blocks.. which is quite frustrating

as an alternative effort I focused on the concept of constant speed.. so concentrating on forces applied to the the lower block

[tex] \Sigma F_x = 0 = F - f_{LT} = F - \mu_{LT} F_N [/tex]

[tex]\Sigma F_y = - W_u - W_L + F_N = 0 [/tex]

[tex]F_N = (m_u + m_l)g [/tex]

[tex]F = \mu_{LT}(m_u + m_l)g [/tex]

entering my numbers.. I get 28.2 N .. which .. according to WebAssign .. is wrong .. so any hints?

when deciphering my subscripts .. U is upper box, L is lower box, T is table

Thanks in advance for any help,

~Phoenix9

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# Two Blocks, Springs, and Friction

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