Aspring 20cmlong is streatched to 25cm by a load of 50Nwhat will be itslength when streatched by 100Nassuming that the..
50 N/25 cm = 100 N/?? cm
50 cm stretch + 20 cm = 70 cm
John, are you sure about this?
stretches 5 with 50 N so stretches 10 with 100 N
20 + 10 = 30
k = 50N/(25-20)cm = 10N/cm.
L = 20 + (1cm/10N * 100N) = 30 cm.
To calculate the length of the spring when stretched by 100N, we can use Hooke's law, which states that the force exerted on a spring is directly proportional to the extension of the spring. Hooke's law is mathematically represented as:
F = k * ΔL
Where:
F is the force applied to the spring
k is the spring constant
ΔL is the change in length or extension of the spring
To find the spring constant, we can divide the force by the initial extension:
k = F / ΔL
Now, let's find the spring constant using the given data:
Initial length (L0) = 20 cm = 0.2 m
Final length (L) = 25 cm = 0.25 m
Force (F) = 50 N
ΔL = L - L0 = 0.25 m - 0.2 m = 0.05 m
Now, we can calculate the spring constant:
k = F / ΔL = 50 N / 0.05 m = 1000 N/m
Now that we have the spring constant, we can use Hooke's law to find the length of the spring when stretched by 100N:
F = k * ΔL
Let's substitute the known values:
100 N = 1000 N/m * ΔL
Now, rearrange the equation to solve for ΔL:
ΔL = F / k
Substitute the values:
ΔL = 100 N / 1000 N/m = 0.1 m = 10 cm
Finally, to find the length of the spring when stretched by 100N, we add the initial length and the change in length:
Length (L) = L0 + ΔL = 0.2 m + 0.1 m = 0.3 m = 30 cm
Therefore, the length of the spring when stretched by 100N will be 30 cm.