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

1. A 2kg block sits on a 4kg block that is on a frictionless table. The coefficients of friction between the blocks are 0.3 (the static) and 0.2 (the kinetic).

a. what is the max F that can be applied to the 4kg block if the 2kg block is not to slide.

b. If F is half this value, find the acceleration of each block and the force of friction acting on each block.

2. m_{1}=250g is at rest on a plane that makes an angle of 30 degree above the horizontal. The coefficient of kinetic friction between the block and the plane is 0.100. The block is attached to a second block of m_{2}= 200g that hangs freely by a string that passes over a frictionless and massless pulley. When the 2nd block has fallen 30cm, its speed is ???

(a) 83 cm/s

(b)48 cm/s

(c)160 cm/s

(d) 59 cm/s

(e) 72cm/s

2. Relevant equations

F=ma, Fx= coeff * F_{n}

3. The attempt at a solution

So I got part a right which is 17.7N. However, I got part b wrong.

1/2 of F= 8.85N.

F_{s,2}= coeff * F_{n}= 0.3.2.9.81= 5.9 N Is this right ??

I got marked wrong on this one but cannot figure out why.

2. So I draw FBD and wrote Newton's 2nd law for each block and get

m_{2}- F_{n}*coeff of friction -m_{1}g*sin[tex]\theta[/tex] =(m_{1}+m_{2})*a.

So I got a and use V_{f}^{2}= 2a[tex]\Delta[/tex]x

I got around 14...cm/s so I chose c but it's wrong. Can you guys help me ??

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# Homework Help: Find the acceleration of each block

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