Question

A 2.00-kg object is free to slide on a horizontal surface. The object is attached to a spring of spring constant 300 N/m , and the other end of the spring is attached to a wall. The object is pulled in the direction away from the wall until the spring is stretched 50.0 mm from its relaxed position. The object is not released from rest, but is instead given an initial velocity of 1.50 m/s away from the wall. Ignore friction.

Answer 1

Answer:

The total energy of the oscillating system is 2.625 J.

Explanation:

Given that,

Mass of object = 2.00 kg

Spring constant = 300 N/m

stretched spring = 50.0 mm

Initial velocity = 1.50 m/s

Suppose we need to determine the total energy of the oscillating system.

We need to calculate the potential energy

Using formula of potential energy

$P.E=\dfrac{1}{2}kx^2$

Put the value into the formula

$P.E=\dfrac{1}{2}\times300\times(50.0\times10^{-3})^2$

$P.E=0.375\ J$

We need to calculate the initial kinetic energy

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times2.00\times(1.50)^2$

$K.E=2.25\ J$

We need to calculate the total energy of the oscillating system

Using formula of total energy

$T.E=K.E+P.E$

Put the value into the formula

$T.E=2.25+0.375$

$T.E=2.625\ J$

Hence, The total energy of the oscillating system is 2.625 J.

Answer 2

Answer:

given,

mass of the object = 2 Kg

spring constant = 300 N/m

distance of stretched = 50 mm = 0.05 m

to determine the total energy of the oscillating system

Potential energy of the spring

= $\dfrac{1}{2}kx^2$

= $\dfrac{1}{2}\times 300 \times 0.05^2$

= 0.375 J

kinetic energy

= $\dfrac{1}{2}mv^2$

= $\dfrac{1}{2}\times 2\times 1.50^2$

= 2.25 J

total energy of the system

Total Energy = PE + KE

Total Energy = 0.375+ 2.25

                     = 2.625 J