The force E needed to make a machine pull a load is partly constant and partly varies as the load L to be pulled.For a load of 20kg the force needed is 14N while 20N is needed for a load of 30kg find: a)The formula connecting E and L. b)E when L=50kg
straight application of a linear function.
let E = mL + k, where E is the force, and l is the load, and k is the constant
given :
when L = 20, E = 14
20m + k = 14
when L = 30, E = 20
30m + k = 20
subtract them:
10m = 6
m = 6/10 = .6
back into 20m + k = 14
12 + k = 14
k = 2
so we have: E = .6L + 2 <------- a)
for b), replace the given L
Pls the correct answer
To find the formula connecting E and L, we need to determine the constant part and the varying part of the force.
Let's denote the constant part of the force as C and the varying part as kL, where k is a constant.
Given that the force needed for a load of 20 kg is 14 N, we can set up the equation:
C + k(20) = 14
Similarly, for a load of 30 kg, the force needed is 20 N:
C + k(30) = 20
Simplifying the equations:
C + 20k = 14 --- Equation 1
C + 30k = 20 --- Equation 2
To find C, we can subtract Equation 1 from Equation 2:
(C + 30k) - (C + 20k) = 20 - 14
10k = 6
k = 6/10
k = 0.6
Substituting the value of k in Equation 1:
C + 20(0.6) = 14
C + 12 = 14
C = 14 - 12
C = 2
Therefore, the formula connecting E and L is:
E = 2 + 0.6L
Now, to find E when L = 50 kg, we can substitute L = 50 into the formula:
E = 2 + 0.6(50)
E = 2 + 30
E = 32
So, when L = 50 kg, the force needed, E, is 32 N.
To find the formula connecting the force E and the load L, we'll first assume that the force needed (E) can be written as a sum of two components: a constant component (C) and a variable component (V) that is proportional to the load (L).
a) Formula connecting E and L:
To find the formula, we'll set up an equation using the given information. We know that for a load of 20 kg, the force needed is 14 N, and for a load of 30 kg, the force needed is 20 N.
First, let's write the equation for the force E when the load is 20 kg:
E = C + V * L
14 = C + V * 20 -- (equation 1)
Next, let's write the equation for the force E when the load is 30 kg:
E = C + V * L
20 = C + V * 30 -- (equation 2)
Now, we have a system of two equations. We can solve these equations simultaneously to find the values of C and V.
Subtract equation 1 from equation 2:
20 - 14 = (C + V * 30) - (C + V * 20)
6 = V * 30 - V * 20
6 = V * (30 - 20)
6 = V * 10
V = 6/10
V = 0.6
Now, substitute the value of V back into equation 1 to find C:
14 = C + 0.6 * 20
14 = C + 12
C = 14 - 12
C = 2
Therefore, the formula connecting E and L is:
E = 2 + 0.6 * L
b) Finding E when L = 50 kg:
Using the formula from part (a), substitute L = 50:
E = 2 + 0.6 * 50
E = 2 + 30
E = 32
Therefore, when L = 50 kg, the force E needed is 32 N.