1. The problem statement, all variables and given/known data A spring of length 0.80 m rests along a frictionless 30° incline (a). A 2.6 kg mass, at rest against the end of the spring, compresses the spring by 0.10 m. (b) http://img64.imageshack.us/img64/5889/physics.jpg [Broken] The mass is pushed down, compressing the spring an additional 0.60 m, and then released. If the incline is 2.0 m long, determine how far beyond the rightmost edge of the incline the mass lands. The spring constant is 127.4 2. Relevant equations Us=1/2kx² v²=v0²+2ax v=v0+at x=x0+v0t+1/2at² 3. The attempt at a solution 0.8m, in a 30° angle, on the spring will count as the 0 point. The potential energy of the spring on the block is 1/2kx²=Us so (1/2)(127.4)(0.1+0.6)²=31.213J Ki+Ui=Kf+Uf 0+31.213=Kf+0 K=1/2mv² 31.213=1/2(2.6)v² 4.9=v http://img691.imageshack.us/img691/5889/physics.jpg [Broken] Diagram of forces show gravity acts against it so mgsin30 (2.6)(-9.8)sin30=-12.74N F=ma -12.74=2.6a -4.9=a v²=v0²+2a(x-x0) v²=4.9²+2(-4.9)(1.9) 2.32=v So the block leaves the ramp at ~2.32m/s, 30° above the horizontal. I use the time it takes for the block to reach its peak height and time to hit the ground and multiply that with the velocity it moves rightwards. v=v0+at 0=2.32sin30+-9.8t 0.12=t1 In this time the block moves up so its final height is x=x0+v0t+1/2at² x=0+(2.32sin30)0.12+(1/2)(-9.8)(0.12)² x=0.06875+1 (original height of right edge of ramp) x=1.06875 Next is the time it takes to land, x=x0+v0t+1/2at² 1.06875=0+0+(1/2)(9.8)t² 0.467=t2 t1+t2=0.585 Finally the distance it goes away from the edge is velocity times time so x=vt x=(2.32cos30)(0.585) x=1.1771557879568892088679537911192 Is this right or are there something wrong with the way I approached the problem?