#4. A 10.0-kg block is released from point A as shown. The track is frictionless

Question

#4. A 10.0-kg block is released from point A as shown. The track is frictionless except for the portion between points B and C, which has a length of 8.00 m. The block travels down the track, hits a spring of force constant 250 N/m, and compresses the spring, and coming to rest momentarily. If the coefficient of kinetic friction between the block and the rough surface between B and C. is 0.13, Detemine: a) The distance that the spring is compressed. b) How far below the point A will the block reach after it leaves the spring for the first time?

Explanation / Answer

velocity of block when it reaches the ground

mgh = 0.5mv^2

v = sqrt(2gh) = sqrt(2*9.8*3) = 7.69 m/s

velcoity of block after it crosses the rough surface

v^2 - u^2 = 2*-(uk*g) *S

v^2 - 7.69^2 = 2*(-0.13*9.8*8)

v = 6.2 m/s

the spring compresses by

0.5kx^2 = 0.5mv^2

250*x^2 = 10*6.2^2

x = 1.24 m

velocity when the block retreates

v^2 - u^2 = 2*(-uk*g)*S

v^2 - 6.2^2 = 2*(-0.13*9.8)*8

v = 4.25 m/s

the block will rise to a height h = v^2 / 2g = 4.25^2 / (2*9.8) = 0.92 m

a) 1.24 m

b) 0.92 m

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