In the figure, a 3.8 kg block is accelerated from rest by a compressed spring of
ID: 2139500 • Letter: I
Question
In the figure, a 3.8 kg block is accelerated from rest by a compressed spring of spring constant 640 N/m. The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction ?k = 0.280. The frictional force stops the block in distance D = 8.0 m. What is the original compression distance of the spring?
In the figure, a 3.8 kg block is accelerated from rest by a compressed spring of spring constant 640 N/m. The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction ?k = 0.280. The frictional force stops the block in distance D = 8.0 m. What is the original compression distance of the spring?Explanation / Answer
gain in kinetic energy by the block=work done by the spring force ---(1)
after leaving the spring at relaxed position ,it travels over the horizontal floor
loss in the kinetic energy by the block = work done by the frictional force --(2)
so,from (1) and (2)
work done by the spring force =- work done by the frictional force
let the initial compression in the spring be x m
force equation in the vertical direction,N-mg=0
=>N=mg
=>N=3.8*9.8=37.24 N
frictional force,f=0.280*37.24=10.43 N
=> (0.5)*k(x^2-0^2) = -(10.43*D*cos (180))
=>(0.5)*640 *x^2=10.43*8.0
=>x=0.5106 m
=>x=51.06 cm
therefore, the original compression is 51.06 cm
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