A block is pushed against the spring with a spring constant 11kN/m located on th

Question

A block is pushed against the spring with a spring constant 11kN/m located on the (left-hand side of the track) and compress the spring a distance of 4.8 cm from its equilibrium position (as shown in figure below) The block starts at rest,is accelerated by the compressed spring, in the slides across a frictionless track except for a small rough area on the horizontal section of the track(shown in green). it leaves the track flies through the air and subsequently strikes the ground. Acceleration due to gravity is not 9.8 m/s^2. a) what is the speed "v" of the block when it leaves the track? answer in units of m/s. b) what is the horizontal distance "x" the block travels in the air? answer in units of m. 11 kN/m 482 g -t t i 4.8 1.5111

Explanation / Answer

(A) Applying work - energy theorem,

Work done by spring + work done by friction = change in KE

(11000)(0.048^2)/2 + (-0.2 x 0.482 x 9.81 x 1.5) = 0.482(v^2/2 - 0 )

v = 6.83 m/s ....Ans

(B) in vertical,

yf- yi = v0y t + ay t^2 / 2

- 1.9 = 0 - 9.81 t^2 /2

t = 0.622 sec

delta(x) = v t = 4.25 m

(C) v^2 - 6.83^2 = 2(-9.81)(-1.9)

v = 9.16 m/s

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