A spring of natural length 1.5m is extended 0.005m by a force of 0.8Newton what will it length be when the applied force is 3.2Newton
k = 0.8N/0.0050m = 160 N/m. L = 1.5 + 3.5N*1m/160N. = 1.5219 m.
(Hope this helps and also let me know if you can't understand this)
I don't understand it
Pls 🙏🙏 explain better
0.8 divided by 0.005 is 160
1.5+3.5N is 5
5x1N is 5
5 divided by 160 is 1.5219
Ajisefinni, I gtg, I was just letting you know, just in case you were going to ask anymore questions, bye.
To determine the length of the spring when the applied force is 3.2 Newton, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its natural length.
The equation for Hooke's Law is as follows:
F = k * x
Where:
F is the force applied to the spring
k is the spring constant
x is the displacement of the spring from its natural length
To find the spring constant, we need to divide the force applied by the displacement of the spring at that force.
First, let's calculate the spring constant (k) using the given values:
k = F / x
k = 0.8 N / 0.005 m
k = 160 N/m
Now that we have the spring constant, we can find the final length of the spring using the new force:
F = k * x
Let's plug in the values:
3.2 N = 160 N/m * x
We can rearrange the equation to solve for x:
x = F / k
x = 3.2 N / 160 N/m
x = 0.02 m
Now we can find the final length of the spring by adding the displacement to the natural length:
Final length = natural length + displacement
Final length = 1.5 m + 0.02 m
Final length = 1.52 m
Therefore, when the applied force is 3.2 Newton, the length of the spring will be 1.52 meters.
3.2 is four times 0.8
so it will be extended four times as far
0.020
plus original length 1.5
= 1.520
3.2 Newtons Shark, not 3.5