Question

when a 4.60-kg object is hung vertically on a certain light spring that obeys hooke's law, the spring stretches 2.80 cm.(a) if the 4.60-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?.91 correct: your answer is correct. cm(b) how much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?1.288 correct: your answer is correct. j

Answer 1

To calculate the spring stretch when a 1.50-kg block is hung on the spring, we can use Hooke's Law. First, we calculate the spring constant based on the stretch of the previous object. Then, we use the calculated spring constant to find the stretch of the spring when the 1.50-kg block is hung.

Step-by-step explanation:

To calculate the spring stretch when a 1.50-kg block is hung on the spring, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In this case, we can use the equation F = kx, where F is the force, k is the spring constant, and x is the displacement.

Given that the displacement of the 4.60-kg object is 2.80 cm, we can rearrange the equation to solve for k:

k = F / x

= (m * g) / x

= (1.50 kg * 9.8 m/s^2) / (2.80 cm)

Using this calculated spring constant, we can now find the stretch of the spring when a 1.50-kg block is hung on it:

x = F / k

= (m * g) / k

= (1.50 kg * 9.8 m/s^2) / (k)

Substituting the given values, we can calculate the spring stretch:

x = (1.50 kg * 9.8 m/s^2) / (0.091 N/m)

= 16.07 cm