21 and 22...the answer to the first problem (part 1) is 3.46 m/s A block is push

Question

21 and 22...the answer to the first problem (part 1) is 3.46 m/s

A block is pushed against the spring with spring constant 4.3 kN / m (located on the left-hand side of the track) and compresses the spring a distance 5.4 cm from its equilibrium position (as shown in the figure below). 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 (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s2. What is the speed nu of the block when it leaves the track? Answer in units of m/s What is the horizontal distance x the block travels in the air? Answer in units of m What is the total speed of the block when it hits the ground? Answer in units of m/s

Explanation / Answer

Part 1

The energy the spring stored is    E = kx2/2 = 4300*0.0542/2 = 6.2694 J

The work done by the friction force is

W = FS = mg *S = 0.5*0.576*9.8*1 = 2.8224 J

The net kinetic energy of the block leaves the track is E = E-W = 3.447 J

From E = mv2/2       =>   v = (2E/m) = 3.46 m/s

Part 2

The horizontal velocity does not change. The time in the air t is

t = (2h/g) = (2*2.4/9.8) = 0.6998 s

The distance is than x = vt = 3.46*0.6998 = 2.42 m

Part 3

From conservation of energy, we have

E+mgh = mv2/2    3.447+0.576*9.8*2.4 =0.576*v2/2     =>

v = 7.68 m/s when the block hits the ground.

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