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Yakvenalex [24]

4 years ago

6

A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The

block then slides to the right on a horizontal frictionless surface and then up a frictionless incline.
Find the maximum height that the block reaches if the incline is 19.5 ∘ . All surfaces are frictionless, the spring constant is 1300 N/m and the initial spring compression is 26.4cm.

1 answer:

Alex787 [66]4 years ago

4 0

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

$\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2$

$Kx^2=mv^2$

$v=\sqrt{\dfrac{kx^2}{m}}$

By putting the values

$v=\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{1300\times 0.264^2}{25}}$

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

$V_f^2=V_i^2-2gh$

By putting the values

$0^2=1.9^2-2\times 10\times h$

h=0.1805 m

h=18.05 cm

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