In the figure below, a 4.3 kg block is accelerated from rest by a compressed spr

Question

In the figure below, a 4.3 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 = 0.25. The frictional force stops the block in distance 8.7 m. (a) What is the increase in the thermal kinetic energy of the block floor system? J (b) What was the maximum kinetic energy of the block? J (c) What was the original compression distance of the spring? m

Explanation / Answer

Given that

A block of mass (m) =4.3kg

Spring constant (k) =640N/m

The coefficient of kinetic friction k = 0.25.

The frictional force stops the block in distance (d) = 8.7 m

a)

The increase in thermal kinetic energy is due to friction is given by

W =F.d =ukmgd =0.25*4.3*9.81*8.7 =91.748J

b)

The maximum kinetic energy is given by

DletaKE =KEfinal -KEinital

Now KEfinal =0

KE inital =Workdone = 91.748J

c)

We know that

KE =(1/2)kx2

Then x=Sqrt(2KE/k) =Sqrt(2*91.748J/640N/m) =0.535m

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