A block with mass m 1 = 8.5 kg is on an incline with an angle = 32° with respect

Question

A block with mass m1 = 8.5 kg is on an incline with an angle = 32° with respect to the horizontal. For the first question there is no friction, but for the rest of this problem the coefficients of friction are: k = 0.28 and s = 0.308.

1)

When there is no friction, what is the magnitude of the acceleration of the block?

2)

Now with friction, what is the magnitude of the acceleration of the block after it begins to slide down the plane?

3)

To keep the mass from accelerating, a spring is attached. What is the minimum spring constant of the spring to keep the block from sliding if it extends x = 0.12 m from its unstretched length.

4)

Now a new block with mass m2 = 15.3 kg is attached to the first block (above the first block on the incline). The new block is made of a different material and has a greater coefficient of static friction. What minimum value for the coefficient of static friction is needed between the new block and the plane to keep the system from accelerating?

Explanation / Answer

mass of the block m1= 8.5 kg ,inclined ar an angle = 32° wrt horizontal, k = 0.28 and s = 0.308.

1)
   The magnitude of the acceleration of the block when there is no friction is

       mg sin = ma ==> a = g sin = 9.8* sin 32 = 5.2 m/s2

2)     With friction, the magnitude of the acceleration of the block after it begins to slide down the plane
      

   limiting friction forcce F = s *mg cos= 0.308*8.5*9.8 cos 32= 21.76 N

now sliding frictional force is = 21.76*0.28/0.308 = 19.78 N

   resulatant force down the plane is = mg sin - 19.78 = 8.5*9.8sin32 - 19.78 = 24.36 N

        F = ma ==. a= F/m = 24.36/8.5 = 2.86 m/s2

3)   x= 0.12 m
to keep the block from sliding without acceleration Ff < or equal to mg cos*s

the final forces can be written as kx + Ff - mg sin =0
                   Ff = mg sin -kx

now    mg sin -kx (mg)coss

   k (mg)(sin-coss)/x
   k = 8.5*9.8(sin32-0.308cos32)/0.12
          = 186.536 N/m

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