Question

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
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

Answer 1

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