A block of mass m = 0.53 kg attached to a spring with force constant 119 N/m is free to move on a frictionless, horizontal surface as in the figure below. The block is released from rest after the spring is stretched a distance A = 0.13 m. (Indicate the direction with the sign of your answer. Assume that the positive direction is to the right.)
The left end of a horizontal spring is attached to a vertical wall, and the right end is attached to a block of mass m which can slide left and right on a horizontal surface. The spring is stretched from its equilibrium position by a distance A
(a) At that instant, find the force on the block. N
(b) At that instant, find its acceleration. m/s2
2)A block of mass m = 2.00 kg is attached to a spring of force constant k = 5.80 ✕ 102 N/m that lies on a horizontal frictionless surface as shown in the figure below. The block is pulled to a position xi = 4.80 cm to the right of equilibrium and released from rest. The left end of a horizontal spring of force constant k is attached to a vertical wall. The right end is attached to a block of mass m, which rests on a horizontal floor. The equilibrium position of the spring is labeled x = 0. The block has been moved to the right to a position labeled x = xi, so that the spring is stretched.
(a) Find the work required to stretch the spring. J
(b) Find the speed the block has as it passes through equilibrium. m/s
1) a) F = - k*x = - 119*0.3 = - 35.7 N
b) F = m*a = - k*x
0.53*a = - 119*0.3
a = - 35.7/0.53
a = - 67.358 m/s^2
Negative sign shows that they are in opposite direction to the applied force.
2) a) W = 0.5*k*x^2
W = 0.5*580*4.8*4.8*10^-2
W = 66.816*10^-2 Joules.
b) now, there is no frictional force hence, kinetic energy at equilibrium is equal to potential energy at equilibrium. So,
K. E = P. E
0.5*m*v^2 = 0.5*k*x^2
v^2 = 580*4.8*4.8*10^-4/2
v = 81.74*10^-2 cm/s
v = 0.8174 m/s
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