A body of mass 100.0g is attached to a hanging spring whose force constant is 10.0 kN/m. The body is lifted until the spring is in its unstretched state. The body is then released. Using the law of conservation of energy,

calculate the speed of the body when it strikes a table 15.0 cm below the release point.

Write an energy equation:

Kinetic energy at impact + elasticPE + graviatationPE=0
where gains in energy are positive.

1/2 mv^2 + 1/2 kx^2 +mg(-x+=0

check my thinking.

Your thinking is correct. The energy equation you have written takes into account the conservation of energy. Let's break down each term in the equation.

1/2 mv^2 represents the kinetic energy of the body at impact. This term accounts for the energy associated with the body's motion.

1/2 kx^2 represents the elastic potential energy stored in the spring. This term accounts for the energy stored in the spring as it is compressed or stretched.

mg(-x) represents the gravitational potential energy of the body. In this equation, we are taking the negative of the gravitational potential energy because we are measuring the distance below the release point.

Setting the sum of all these energies equal to zero ensures that the total mechanical energy of the system (kinetic energy + elastic potential energy + gravitational potential energy) is conserved.

Now let's solve the equation to find the speed of the body when it strikes the table.

1/2 mv^2 + 1/2 kx^2 + mg(-x) = 0

Since the spring is initially in its unstretched state, the value of x is 0. So we have:

1/2 mv^2 + 0 + 0 = 0

1/2 mv^2 = 0

Solving for v, we have:

v^2 = 0

v = 0

According to this equation, the speed of the body when it strikes the table is 0. This result suggests that the body does not reach the table, or it stops just before reaching it. Another possibility is that there might be some external factors (such as air resistance) that are not considered in this simplified analysis.