A 1-kg block slides along frictionless surface XY with a velocity of v = 10 m/s.
ID: 2038529 • Letter: A
Question
A 1-kg block slides along frictionless surface XY with a velocity of v = 10 m/s. It then moves along a surface YZ with length 10 m, and uk = 0.2 until hitting an undeformed spring whose k = 1000 N/m.What is the block’s velocity just before it hits the spring?
What will be the maximum compression of the spring?
After leaving the spring, will the block reach surface XY?
If yes, compute for the velocity of the block once it reaches surface XY. If not, how far from B will tbe block be once it stop? A 1-kg block slides along frictionless surface XY with a velocity of v = 10 m/s. It then moves along a surface YZ with length 10 m, and uk = 0.2 until hitting an undeformed spring whose k = 1000 N/m.
What is the block’s velocity just before it hits the spring?
What will be the maximum compression of the spring?
After leaving the spring, will the block reach surface XY?
If yes, compute for the velocity of the block once it reaches surface XY. If not, how far from B will tbe block be once it stop? A 1-kg block slides along frictionless surface XY with a velocity of v = 10 m/s. It then moves along a surface YZ with length 10 m, and uk = 0.2 until hitting an undeformed spring whose k = 1000 N/m.
What is the block’s velocity just before it hits the spring?
What will be the maximum compression of the spring?
After leaving the spring, will the block reach surface XY?
If yes, compute for the velocity of the block once it reaches surface XY. If not, how far from B will tbe block be once it stop?
What is the block’s velocity just before it hits the spring?
What will be the maximum compression of the spring?
After leaving the spring, will the block reach surface XY?
If yes, compute for the velocity of the block once it reaches surface XY. If not, how far from B will tbe block be once it stop?
Explanation / Answer
Given,
m = 1 kg ; Vxy = 10 m/s ; YZ = 10 m ; uk = 0.2 ; k = 1000 N/m
from conservation of energy
KEi - Wf = KEf
where, KEi and KEf are the intial and final kinetic energies.
KEf = 1/2 m v^2 - uk m g d
KEf = 0.5 x 1 x 10^2 - 0.2 x 1 x 9.81 x 10 = 30.38 J
1/2 m vf^2 = 30.38 J
vf = sqrt (2 x 30.38/1) = 7.79 m/s
Hence, vf = 7.79 m/s
Again from conservation of energy
1/2 m vf^2 = 1/2 k x^2
x = v sqrt(m/k) = 7.79 sqrt(1/1000) = 0.25 m
Hence, x = 0.25 m
again it will come across the frictional surface YZ, but will be left with energy to enter the region XY. So
E(left) = KEf - Wf
E(left) = 1/2 x 1 x 7.79^2 - 0.2 x 1 x 9.81 x 10 = 10.72 J
vf' = sqrt (2 x 10.72/1) = 4.63 m/s
Hence, vf' = 4.63 m
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