A 1.60 block slides with a speed of 0.850 m, on a frictionless horizontal surfac
ID: 1536891 • Letter: A
Question
A 1.60 block slides with a speed of 0.850 m, on a frictionless horizontal surface until it encounters a spring with a force constant of 671 The block cones to rest after compressing the spring 4.15 c, . Find the spring potential energy, the kinetic energy of the block, and the total mechanical energy of the system, for compressions of 0 cm. Enter your answers numerically separated by commas. Find the spring potential energy, the kinetic energy of the block, and the total mechanical energy of the system, for compressions of 1.00 cm Enter your answers numerically separated by commas. Find the spring potential energy, the kinetic energy of the block, and the total mechanical energy of the system, for compressions of 2.00 cm Enter your answers numerically separated by commas. Find the spring potential energy, the kinetic energy of the block, and the total mechanical energy of the system, for compressions of 3.00 cm Enter your answers numerically separated by commas. Find the spring potential energy, the kinetic energy of the block, and the total mechanical energy of the system, for compressions of 4.00 cm Enter your answers numerically separated by commas.Explanation / Answer
Given
mass of block m =1.6 kg, speed of block before encontering the spring is v = 0.850 m/s
the surface is friction less
spring constant k = 671 N/m
block comes to rest after compressing the spring dx = 4.15 cm = 0.0415 m
we know that the mechanical energy of the given system is
sum of k.e of block and p.e of the spring which is cosntant
that is E = 0.5 mv^2+ 0.5 kx^2
Part A
when the block is at x= 0 cm , the p.e = 0 J, k.e = 0.5*1.6*0.850^2 J = 0.578 J
so the mechanical energy = 0 +0.578 J = 0.578 J
Part B
when the block compressed the spring about 1 cm then the , p.e = 0.5*671*0.01^2 J = 0.03355 J
total mechanical energy is constant that is E = p.e+k.e ==> k.e = E - p.E = 0.578 -0.03355 J = 0.54445 J
Part C
when the block compressed the spring about 2 cm then the , p.e = 0.5*671*0.02^2 J = 0.1342 J
total mechanical energy is constant that is E = p.e+k.e ==> k.e = E - p.E = 0.578 -0.1342 J = 0.4438 J
Part D
when the block compressed the spring about 3 cm then the , p.e = 0.5*671*0.03^2 J = 0.30195 J
total mechanical energy is constant that is E = p.e+k.e ==> k.e = E - p.E = 0.578 - 0.30195 J = 0.27605 J
Part E
when the block compressed the spring about 4 cm then the , p.e = 0.5*671*0.04^2 J = 0.5368 J
total mechanical energy is constant that is E = p.e+k.e ==> k.e = E - p.E = 0.578 - 0.5368J = 0.0412 J
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