A 1.5-kg block slides at rest starts sliding down a snow-covered hill Point A, w
ID: 1569391 • Letter: A
Question
A 1.5-kg block slides at rest starts sliding down a snow-covered hill Point A, which has an altitude of 10 m. There is no friction on hill. After leaving the hill at point B, it travels horizontally toward a mass-less spring with force constant of 200 N/m. While travelling, it encounters a 20-m patch of rough surface CD where the coefficient of kinetic friction is 0.15. (a) What is the speed of the block when it reaches point B? (b) Find the work dissipated by friction as the block reaches point D. (c) How far will the block compress the spring? Liquid water pressurized to 1.5 times 10^5 Pa is flowing at 25.0 m/s in a horizontal pipe with diameter of 4-cm, which expands to a diameter of 10-cm. (a) Determine the flow rate of water in the pipe. (b) What is the velocity of the water after the expansion (V_2)? (c) Compute the pressure after the expansion joint. (P_2).Explanation / Answer
1. Using conservation of energy at A and B,
mgh = mv^2/2
v= sqrt( 2 g h ) = sqrt( 2 x 9.8 x 10 ) = 14 m/s. (a)
Work done by force of friction= (mu)mgx
= 0.15 x 1.5 x 9.8 x 20 = 44.1 J (b) have been dissipated.
So, now we have k x^2 / 2 = E
E is the leftover energy of the object, given by mgh - 44.1 = 147 - 44.1 = 102.9 J
So, distance compressed for the spring= 1.014 m approx. (c)
2. Flow rate is the rate of volume of water flowing out which is Area x velocity , for that particular point.
25 x pi x 4 x 10^-4 m^3/ s = 3.14 x 10^-2 m^3/ s (a)
Simply use the equation of continuity, A1 v1 = A2 v2 and you'll get v2 (I've calculated A1 v1 in the previous step, already.) (b)
Using Bernoulli's equation,
P1 + pgh1 + pv1^2 / 2 = P2 + pgh2 + pv2^2 / 2 {p is the density of liquid while P is the pressure}
As the setup is at the same height, so the term pgh cancels out, on both the sides of the equation.
Since we are already given P1 and have calculated rest of the parameters, P2 can be calculated:
P1 + pv1^2 / 2 = P2 + pv2^2 / 2 (c).
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