A block is pushed against the spring with spring constant k and compresses the s

Question

A block is pushed against the spring with spring constant k and compresses the spring a distance 4.6 cm from its equilibrium position. The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless track except for a small rough area on a horizontal section of the track. It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.81 m/s2. A) What is the spring constant k? b) What is the speed v of the block when it leaves the track? c) What is the total speed of the block when it hits the ground? Distance compresses: 4.6cm; Mass of the block: 497g; Height: 2.3m; (mu): .5; Distance of the friction: 1.1m;Distance as it leaves the track and hit the ground: 2.51m; Gravity: 9.8 m/s^2.

Explanation / Answer

Time of flight T= sq rt (2h/g) = sq rt 4/9.81=0.6386 s Horizontal velocity v= range / time of flight=4.94 /0.6386 v=7.736 m/s Suppose the block leaves the spring with speed u , Retardation due to friction a=(mu)g= -0.2*9.81= -1.962 m/s^2 Distance s = 1m v^2 -u^2 =2as gives u = sq rt [v^2 - 2as ] u = 7.98 m/s (1/2) kx^2=(1/2)mu^2 (a) k= m[u/x]^2=0.498 [7.98/0.046]^2=14987 N/m (b) The block leaves the track with speed 7.736 m/s (c) V= sq rt [v^2 +2gh]= sq rt [59.845 + 39.24 ]=9.95 m/s the total speed of the block when it hits the ground is 9.95 m/s * 5 years ago

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