A small block of mass Af is projected upward with a speed A massless spring of s
ID: 3281464 • Letter: A
Question
A small block of mass Af is projected upward with a speed A massless spring of stiffness constant A hangs from the ceiling Its bottom is a height h above the grount Let AM-2.5 kg-30ms,200 N/m and h -24 A) Find the total mechanical energy ) 2800b 1840, 1125 ) 62s D) Find the speed of the block when the block hits the speing a) 27.5m's b) 20.7 m/s ) 12.4 m's, d) 33.6ms C) The block sticks to the spring Find the maximum compression D) When the mass reaches its lowest print find the height above the ground. h" E) If it didn't stick to the spring find the speed " of the mass just before it lands a) 2.20m, b) 1.15 m )4.28m d) 3.22 m a) 22.7m. b) 19.4m e 21.6m. d) 20.5m a) 35 mis, b) 25 S )40 m's d) 30 m's Problem 2 A force Fis applied to a block of mass M which is attached to a spring with spring constant A. The block slides a distance D along a table with coefficient of sliding friction 0.0 kg, k-35 N/rn. D .7 m, and Pk-0.35 Let F-55 N, M A) Find the work done by the force F B) Find the change in the potential energy of the spring C) D) Find the work done by the friction force on the block E) Find the final velocity of the block ) 93.5, b) 114J. c)130, d) 75J a) 90.31 b) 80Jc) 79.5J,d) 67.5 a) 2.21 N, b) 1.96 N, c) 2.06 N, d) 3,09 N a) 3.50J, b) -2.94J, e) -4.41J d) -5.87J a) 5.92 /s, b) 7.15 m/s. c) 10.I m's. d) 4.27 m/s Find the rnagnitude of the friction force.Explanation / Answer
A] total mechanical energy = 0.5 mv0^2 = 0.5*2.5*30^2 = 1125 J option C
B] By energy conservation, 0.5mv^2 + mgh = 1125
v = sqrt(2*(1125-mgh)/m)
v = sqrt(2*(1125-2.5*9.8*24)/2.5) = 20.7 m/s. option b is correct.
c] Let it be x, mg(h+x) + 0.5kx^2 = 1125
2.5*9.8*(24+x) + 0.5*200x^2 =1125
solving quadratic equation, x = 2.20 m option A
D]Let the extension be x, mg(h-x) + 0.5kx^2 = 1125
2.5*9.8*(24-x) + 0.5*200x^2 =1125
x = 2.4 m,
so height = 24-2.4 m = 21.6 m option c
E] v =v0= 30 m/s , option d