A spring with a force constant of 120N/m is used to push a 0.27kg block of wood against a wall.
a. find the minimum compression of the spring needed to keep the block from falling, given that the coefficient of static friction between the block and the wall is 0.46.
b. does your answer to part A change if the mass of the block of wood is doubled?
Please show all your work 10 points to best answer.
3 Answers
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For the block to stay there, the force of friction ‘f’ should be equal to weight ‘mg’ of the block. For the frictional force, the normal reaction ‘N’ is provided by the spring force ‘F’. The coefficient of friction ‘u’ = 0.46.
A) uN = mg
=> 0.46 x N = 0.27 x 9.8
=> N = 5.75 newton.
This should be the normal force on the block by the wall. Thus minimum spring force should be 5.75 newton.
F = k x d (where k = spring constant, d = compression in spring)
=> 5.75 = 120 x d
=> d = 0.048 m
=> d = 4.8 cm
B) of course the answer changes if the mass of block is doubled. For the mass 0.54 kg , the normal force that is needed is 11.4 newton. And the compression becomes 9.6 cm.
Source(s): My brain is my source. -
So the block doesnt fall, the friciton force (Ffr) has to be at least the weight of the object (mg). Ffr=mg
Ffr=mg=Fn*coefficient of friction
Fn=.mg/.46
The normal force is provided by the wall and depends on how hard you are pushing on the wall with the spring. For springs, F=kx, so F=kx=Fn. mg/.46=120*x x=.049m
(I used 10 for g)
b. Yes, see the equation^
Source(s): ap physicsssssssssss -
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