007 (part 1 of 3) 10.0 points A block is pushed against the spring with spring c

Question

007 (part 1 of 3) 10.0 points A block is pushed against the spring with spring constant 6.5 kN/m (located on the left- hand side of the track) and compresses the spring a distance 4.9 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/s 6.5 kN/m 476 g --Vz p00999.. 4.9 cm 1.1 m What is the speed v of the block when it O leaves the track?

Explanation / Answer

Spring constant, K = 6.5 KN/m = 6500 N/m

Energy stored in the spring as spring energy, P1 = (1/2)*K*x^2 = 0.5*6500*0.049^2 = 7.803 J.

Part 1 –

mass of the block, m = 476 g = 0.476 kg.

Work done against the friction force, Ws = u*m*g*d = 0.3*0.476*9.8*1.1 = 1.54 J

So, the net energy converted into the kinetic energy of the block KE = P1 – Ws = 7.803 – 1.54 = 6.263 J

Suppose ‘v’ is the velocity of the block when it leaves the track.

So, KE = (1/2)*m*v^2

=> 6.263 = 0.5*0.476*v^2

=> v = 5.13 m/s

Part 2 –

Height of the track, h = 1.9 m.

now assume time taken by the block to reach at the ground is t sec.

use the expression –

h = u*t + (1/2)*g*t^2

=> 1.9 = 0 + 0.5*9.81*t^2

=> t^2 = 1.9 / (0.5*9.81) = 0.62 sec.

So, horizontal distance travelled by the block in air, x = v*t = 5.13*0.62 = 3.18 m

Part 3 –

Vertical component of the velocity, Vy = 9.81*0.62 = 6.08 m/s.

So total speed of the block when it hits the ground, V = sqrt[5.13^2 + 6.08^2] = 8.0 m/s.

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