Question

A horizontal spring attached to a wall has a force constant of k = 860 N/m. A block of mass m = 1.60 kg is attached to the spring and rests on a frictionless, horizontal surface as in the figure below. 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. The spring has force constant k. Three positions are labeled along the spring, and the block is pulled to the rightmost position, stretching the spring. The leftmost position, the equilibrium position, is labeled x = 0. The middle position, halfway between the leftmost and rightmost positions, is labeled x = xi⁄2. The rightmost position is labeled x = xi. (a) The block is pulled to a position xi = 5.00 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 5.00 cm from equilibrium. J (b) Find the speed of the block as it passes through the equilibrium position. m/s (c) What is the speed of the block when it is at a position xi/2 = 2.50 cm? m/s

Answer 1

(a) 1.08 J

The elastic potential energy stored in the block at any position x is given by

$U=\frac{1}{2}kx^2$

where

k is the spring constant

x is the displacement relative to the equilibrium position

Here we have

k = 860 N/m

x = 5.00 cm = 0.05 m is the position of the block

Substituting, we find

$U=\frac{1}{2}(860 N/m)(0.05 m)^2=1.08 J$

(b) 1.16 m/s

The total mechanical energy of the spring-mass system is equal to the potential energy found at point (a), because there the system was at its maximum displacement, where the kinetic energy (because the speed is zero).

At the equilibrium position, the mechanical energy is sum of kinetic and potential energy

E = K + U

However, at equilibrium position x = 0, so U = 0. Therefore, the kinetic energy is equal to the total energy found at point (a)

$E=K= \frac{1}{2}mv^2 = 1.08 J$

where

m = 1.60 kg is the mass of the block

v is the speed

Solving for v, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.08 J)}{1.60 kg}}=1.16 m/s$

(c) 1.00 m/s

When the block is at position x = 2.50 cm, the mechanical energy is sum of kinetic and potential energy:

$E=K+U=\frac{1}{2}mv^2 + \frac{1}{2}kx^2$

where

E = 1.08 J is the total mechanical energy

m = 1.60 kg is the mass

v is the speed

k = 860 N/m

x = 2.50 cm = 0.025 m is the displacement

Solving for v, we find

$v = \sqrt{\frac{2E - kx^2}{m}}=\sqrt{\frac{2(1.08 J)-(860 N/m)(0.025 m)^2}{1.60 kg}}=1.00 m/s$