A spring sketch by 6cm when supporting a load of 15N? By how much would it stretch when supporting a load of 5Kg
Please any other way to calculate it
F = k x
15 = k * 6
k = 15/6
5 = (15/6) x
x = 5 * (6/15 ) = 2 cm
Ratio way
15÷5=6÷x
Cross multiplication
15x=30
x=30÷15
x=2
15000kg-6
can yall explain what ur doing step by step in simpler terms
To answer this question, we need to understand Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount it stretches or compresses. The formula for Hooke's Law is:
F = k * x
where F is the force applied to the spring, k is the spring constant, and x is the displacement or stretch/compression of the spring.
To find the stretch of the spring when supporting a load of 5 kg, we first need to determine the force exerted by the load. The force due to gravity is given by:
F = m * g
where m is the mass of the load and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given that the mass is 5 kg, the force exerted by the load can be calculated as:
F = 5 kg * 9.8 m/s² = 49 N
Now, let's find the spring constant. The spring constant reflects the stiffness of the spring and is typically given in N/m. Since we don't have the spring constant directly, we can calculate it using the first measurement provided.
When supporting a load of 15 N, the spring stretched by 6 cm. Using Hooke's Law, we can rearrange the formula to solve for the spring constant:
k = F / x
where F is the force and x is the displacement.
By substituting the given values, we have:
k = 15 N / 0.06 m = 250 N/m
Now that we have the spring constant, we can find the stretch of the spring when supporting a load of 5 kg (49 N). Again using Hooke's Law:
F = k * x
Rearranging the formula to solve for x:
x = F / k
Substituting the known values:
x = 49 N / 250 N/m = 0.196 m = 19.6 cm
Therefore, the spring would stretch by approximately 19.6 cm when supporting a load of 5 kg.
Mg = 5*9.8 = 49 N.
15/6 = 49/x
X = 19.6 cm.