The scale of a certain spring balance reads 1200 N when the spring is stretched 0.1m.
(a) What is the potential energy of the spring when it is stretched 0.1m and when it is stretched 0.5 m?
(b) What is the potential energy when a 50 kg mass hangs from the spring?
To answer these questions, we need to understand the relationship between the potential energy of a spring and its stretch.
The potential energy of a stretched or compressed spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement or stretch from its equilibrium position.
Hooke's Law can be expressed as follows: F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement or stretch of the spring.
(a) To calculate the potential energy of the spring when it is stretched 0.1m and 0.5m, we first need to find the spring constant.
Using the information given in the question, we know that when the spring is stretched 0.1m, the scale reads 1200 N.
Since F = kx, we can rearrange the equation to solve for k: k = F/x.
k = 1200 N / 0.1m
k = 12000 N/m
Now that we know the spring constant, we can calculate the potential energy when the spring is stretched 0.1m and 0.5m using the formula for potential energy of a spring:
PE = (1/2) kx^2
For the 0.1m stretch:
PE = (1/2) * (12000 N/m) * (0.1m)^2
PE = 60 N * m = 60 joules (J)
For the 0.5m stretch:
PE = (1/2) * (12000 N/m) * (0.5m)^2
PE = 1500 N * m = 1500 joules (J)
Therefore, the potential energy of the spring when it is stretched 0.1m is 60 joules (J), and when it is stretched 0.5m, it is 1500 joules (J).
(b) To calculate the potential energy when a 50 kg mass hangs from the spring, we need to first find the stretch of the spring caused by the weight of the mass. We can use the equation F = kx, where F is the weight of the object (mg).
F = (50 kg) * (9.8 m/s^2) = 490 N
Now, using the spring constant we calculated earlier (k = 12000 N/m), we can solve for the displacement x:
490 N = (12000 N/m) * x
x = 490 N / 12000 N/m
x = 0.0408 m
Now that we know the stretch of the spring caused by the 50 kg mass is 0.0408 m, we can calculate the potential energy as:
PE = (1/2) * (12000 N/m) * (0.0408 m)^2
PE = 9.99 N * m = 9.99 joules (J)
Therefore, the potential energy of the spring when a 50 kg mass hangs from it is approximately 9.99 joules (J).