DOC

# problem07_74

By Monica Willis,2014-12-06 03:40
19 views 0
problem07_74

7.74: a) From either energy or force considerations, the speed before the block hits the spring is

2(sincos)vgLk

22(9.80ms)(4.00m)(sin53.1(0.20)cos53.1)

7.30ms.

b) This does require energy considerations; the combined work done by gravity and

21kdfriction is mg(L;d)(sinθcosθ), and the potential energy of the spring is , k2

where d is the maximum compression of the spring. This is a quadratic in d, which can be

written as

k2ddL0. 2mg(sincos)k

124.504mThe factor multiplying is , and use of the quadratic formula gives d

d1.06m. c) The easy thing to do here is to recognize that the presence of the spring determines d, but at the end of the motion the spring has no potential energy, and the distance below the starting point is determined solely by how much energy has been lost to friction. If the block ends up a distance y below the starting point, then the block has

moved a distance down the incline and L;dy up the incline. The magnitude of L;d

mgcosθthe friction force is the same in both directions, , and so the work done by k

(2L;2dy)mgcosθfriction is . This must be equal to the change in gravitational k

potential energy, which is mgysinθ. Equating these and solving for y gives

2cos2kkyL;dL;d()(). ;;sincostankk

Using the value of d found in part (b) and the given values for and gives k

y1.32m.

Report this document

For any questions or suggestions please email
cust-service@docsford.com