1. A block of mass 0.260 kg is placed on top of a light, vertical spring of forc
ID: 1266690 • Letter: 1
Question
1. A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 4 800 N/m and pushed downward so that the spring is compressed by 0.091 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.) 2. Two objects are connected by a light string passing over a light, frictionless pulley as shown in the figure below. The object of mass mi = 6.00 kg Is released from rest at a height h = 5.00 m above the table. (a) (a) Using the isolated system model, determine the speed of the object of mass m2 = 3.00 kg just as the 6.00-kg object hits the table. (b) Find the maximum height above the table to which the 3.00-kg object rises.Explanation / Answer
apply force equation on system
6*g - T = 6a .......and T - 3*g = 3a
so accelaratin = g/3
a)
speed of block = sqrt( 2*a*h ) = sqrt ( 2* 9.81* 5/3) = 5.71 m/sec
B. height (h) = V^2 / 2A = 5.71^2* 3 /(2*2*9.81) = 2.5 m .where A = g- a = 2g/3
maximum height = height + initail height = 5 + 2.5 = 7.5 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.