CATAGORY: Mechanics QUESTION: A block is put on a horizontal table with which it has a coefiicient of friction of 0.25. While on theis table the block is pushed up against a spring such that the spring is compressed and the block is stationary. The block is now released and it slides 60 cm across the table before coming to a stop. Now the same experiment is repeated except that the spring is compressed twice as much as before. What is the distance the block slides across the table in this new case?

Back to subcatalog.

Some interpretation first I assume you mean 60cm AFTER the block clears the spring. When the spring is compressed a distance X the block slides 60cm | | before after | _ _ _ _ _ |_/ \_/ \_/ \-|_|--X--|-----60cm-------|_| ---------------|------|-----------------| 0 The energy put into the block by the spring is .5 k X^2 Where k is the spring constant. The energy drain on the block from friction is Uk Mg(X+.6m) where Uk is the coeficient of sliding friction, M is the mass of the block anfd g is the acceleration due to gravity. From conservation of energy we have (1) Uk Mg (X + .6m) = .5 k X^2 For the new arrangement we have a similar equation except X -> 2X and .6m -> L | | before after | _ _ |/\/\/\|_|-----2X-----|----------L=?-------|_| ---------------|------|---------------------| 0 (2) Uk mg (2X + L) = .5 k (2X)^2 We have Uk =.25 but we dont have M, k, X, or L which we are trying to find. We have only two equations and three unknowns. :-( Unless we make further assumptions we can not go further. Suppose that the energy lost from friction can be neglected while the pring is still pushing. Presumably the distance is so short compared to the length of the free sliding part that we can ignore it. If so then we have (3) Uk M g .6m = .5 k X^2 (4) ---> 1.2m Uk M g/k =X^2 and (5) Uk M g L = .5 k (2X)^2 (6) ---> L = .5 k (2X)^2 / Uk M g Substitution of (4) into (6) gives (7) L = .5 k 4(1.2 Uk M g/k) / Uk M g = .5 x 4 x 1.2m = 2.4m (or four times as far) ************ * * * L = 2.4m * * * Approximate answer ************