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 10 kN/m (located on the left- hand side of the track) and compresses the spring a distance 4.5 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 10kN/m 548 g rz 1.5 m =0.3 4.5 What is the speed leaves the track? of the block when it Answer in units of m/s. 008 (part 2 of 3) 10.0 points What is the horizontal distance travels in the air? the block Answer in units of m 009 (part 3 of 3) 10.0 points What is the total speed of the block when it hits the ground? Answer in units of m/s.

Explanation / Answer

1)

Let : g = 9.81 m/s^2

m = 0.548 kg

k = 10000 N/m

vx = v

d = 0.045 m

l = 1.5 m

u = 0.3

from conservation of mechanical energy,

Ui = Uf + Kf

since vi = 0 m/s

k = 1/2mv^2 and Us = 1/2kd^2

Choosing the point where the block leaves the track as the origin of the coordinate system,

x = vxt

y = 1/2gt^2

since axi = 0 m/s^2 and vyi = 0 m/s.

1/2mvx^2 = 1/2kd^2 - umgl

vx = sqrt[(kd^2 / m) - 2ugl]

vx = sqrt(10000 x 0.045^2) / 0.548 - (2 x 0.3 x 9.81 x 1.5)]

vx = 5.3 m/s

2)

At y = h = 2.4 m (below the jump off height)

h = -1/2gt^2

t = sqrt(-2h/g)

x = vx*t

x =  sqrt[(kd^2 / m) - 2ugl] x  sqrt(-2h/g)

x = 3.70 m

3)

Kf = Ui ,

since vi = 0 m/s hf = 0 m

using, 1/2mvy^2 = mgh

vy = sqrt(-2gh)

vy = sqrt(2 x 9.81 x 2.4)

vy = 6.86 m/s

v = sqrt(vx^2 + vy^2)

v = sqrt(5.30^2 + 6.86^2)

v = 8.55 m/s

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