Problem 4 For this problem, imagine that you have two blocks connected by a spri

Question

Problem 4 For this problem, imagine that you have two blocks connected by a spring and able to slide without friction on an air table. The blocks have an inertia of 4kg each. The spring is initially relaxed and has a spring constant of 20Supose you pull on one block with some constant horizontal force. At first the spring stretches, so the separation between the blocks increases, but once the spring is stretched by 0.1 m the separation stops changing, and the blocks and spring move together as a whole. (a)Based on this information, with what force are you pulling on the block? (b)Now, let's think about as the blocks first started moving, but before the spring reached the constant stretch. Assume that, before they started to move (spring relaxed) they were 20 cm apart. Call the location of the block you are not directly pulling 0 at the start. At some instant after you started pulling with the constant force you found above (time ti), imagine that the block you were pulling had moved 4cm (so it is now at 24 cm) and the following block had moved 2 cm, to 2 cm. Where was the center of mass of the two identical blocks before they started to move? How far has the center of mass moved from the time you started until t1? How far has the point of application of the force moved in this time? What is the work done on the system by the pull? What is the translational kinetic energy of the system? Why are these two numbers different (if they are)?

Explanation / Answer

4. assuming two masses connected by a spring on a frictionless table

mass of each block, m = 4 kg

spring constant, k = 240 N/m

maximum strech, A = 0.1 m

a. let the force on the block be A

then for the block moving without acceleration during the streching of the spring

F = kA = 0.1*240 = 24 N

b. assuming origin to be at the block which is not being pulled directly

initial location of other block , x = 0.2 m

at time t1, locaiton of block 1 , x = 0.0 2m

location of second block x = 0.24 m

a. initial center of mass = 0.2/2 = 0.1 m

b. till time t1, locaiton of center of mass = (0.24 - 0.02)/2 = 0.11 m

movement of center of mass dx = 0.01 m

c. point of applicatino of force has moved 0.04 m in this time

d. work done by the system by the pull = F*0.04 = 24*0.04 = 0.96 J

e. translational kinetic energy = work done on the system - energy stored in the spring

energy stored in the spring = 0.5*k*(0.24 - 0.02 - 0.2)^2

W = 0.5*240*0.02^2 = 0.048 J

so TKE = 0.96 - 0.048 = 0.912 J

f. the two numbvers are different as some of the energy given to the systgem is stored in the spring

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