# Force needed to move blocks against friction

Two blocks of masses 3kg and 5kg are connected by a spring of stiffness k. The coefficient of friction between the blocks and surface is 0.2 . Find the minimum constant horizontal force F to be applied to 3kg block in order to just slide the 5kg block.

My work : For 5kg block to just slide, the spring force should be equal to friction experienced by it which is equal to 10N. Now the external force F applied on 3kg block should be greater than or equal to the spring force + friction due to surface which is equal to 10N + 6N = 16N . So my result is that the minimum force should be equal to F = 16N. But this is not the answer. I want to know where am I wrong. Ans - F =11 N

• I've started and erased this comment three times. I don't quite understand the problem. It seems one has to interpret "just slide" as "just exceed the limit of static friction". But I don't know what to make of the friction on the 3 kg block. One would have to first exceed the static friction, compress the spring, then reduce the applied force until it was below the limit of static friction again. Perhaps the question is: what is the smallest force that will keep both objects just below the static limit once the spring is compressed. (I haven't worked that problem out.) Not sure !! Aug 16 '16 at 20:11
• Trying very hard too here and not getting anywhere either.
– Gert
Aug 16 '16 at 20:31
• I got 16 N too. It's possible that there is a typo or error in the answer. I've seen this before (i.e., errors in a book's answer key). Aug 16 '16 at 21:34
• Isn't there a diagram with the problem? The question does not say that the "surface" is horizontal. Aug 17 '16 at 12:40
• No, there's no diagram. Aug 18 '16 at 17:45

For 3kg block:- By work- energy theorem, Work by external force + work by non-conservative force = change in K.E + change in P.E Let required extension of spring be x metres So, Fx - mu(=0.2) × 3 × g(=10) × x = change in P.E + change in K.E As from the equation, eveything except change in K.E is fixed , so for minimum value of F , K.E should be equal to zero. So assuming change in K.E. is very negligible Fx - 0.2 × 3 × 10 × x = 0.5kx^2 + 0 F - 0.2 × 3 × 10 = 0.5kx Since to move the 5 kg block, kx minimum magnitude is 10 N. So- F - 6 = 5 So, F = 11 N

• Can you please reformat this using MathJax? Here’s a MathJax tutorial Aug 18 '16 at 19:10
• Yeah, In some hours Aug 21 '16 at 14:03

The force causes acceleration on $m_1$ (right hand block) and work is done on the spring: $$(F-\mu m_1g)x=\frac12 kx^2$$ The maximum spring extension before $m_2$ starts to move is: $$x_{max}=2\Big(\frac{F-\mu m_1g}{k}\Big)$$ This causes the tension in the spring to reach a maximum: $$T_{max}=kx_{max}=2(F-\mu m_1g)$$ Which has to exceed the friction provided by $m_2$, to cause motion of $m_2$: $$2(F-\mu m_1g)>\mu m_2g$$ $$\implies F>\mu(\frac{m_2g}{2}+m_2g)$$

With the provided numerical values this gives:

$$F>11\:\mathrm{N}$$

If this condition isn't met then $m_1$ will simply come to a halt and $m_2$ won't start moving.

• Gert, I'm not claiming that you or I are infallible. However, this seems to be a relatively straight-forward problem. So ... unless the OP left out something in the problem statement, I'm willing to go with the 16N answer. And one more thing ... the problem statement is somewhat ambiguous, which doesn't help when trying to work the problem. Aug 17 '16 at 1:30
• I didn't assume any acceleration. As I previously stated, the problem statement needs some work. Aug 17 '16 at 3:08