Answer :
Answer:
12 J
Explanation:
From Hook's law,
E = 1/2ke² ................. Equation 1
Where E = Work done in stretching the spring, k = spring constant, e = extension.
make k the subject of the equation
k = (2E/e²).................. Equation 2
Given: e = 10 cm = 0.1 m, E = 4 J.
Substitute into equation 2
k = 2(4)/(0.1²)
k = 8/0.01
k = 800 N/m.
When the spring stretch by an additional 10 cm
E' = 1/2ke'²
e' = 10+10 = 20 cm = 0.2 m.
E' = 1/2(800)(0.2²)
E' = 400(0.04)
E' = 16 J.
ΔE = E'-E
Where ΔE = additional work.
ΔE = 16-4
ΔE = 12 J
Answer:
12J
Explanation:
The work done (W) is stretching a spring that obeys Hooke's law is given by;
W = [tex]\frac{1}{2}[/tex] k x² -------------(i)
Where;
k = the spring constant measured in N/m
x = extension of the spring.
From the question;
When the spring is stretched (extended) by 10cm, the workdone is 4J.
This implies that;
At x = 10cm = 0.1m, W = 4J
Substitute these values into equation (i) to get the spring constant (k) of the spring as follows;
4 = [tex]\frac{1}{2}[/tex] * k (0.1)²
4 = [tex]\frac{1}{2}[/tex] * k (0.01)
4 = 0.005k
k = 4 / 0.005
k = 800N/m
Therefore, the spring constant of the spring is 800N/m
Now, let's get the work done in stretching the spring an additional 10cm.
This means that the total extension X, is the sum of the initial extension (10cm) and the current extension (10cm).
i.e
X = 10cm + 10cm = 20cm = 0.2m
Substitute the value of X for x in equation (i) to get the work done as follows;
W = [tex]\frac{1}{2}[/tex] k X²
W = [tex]\frac{1}{2}[/tex] k (0.2)²
W = [tex]\frac{1}{2}[/tex] k (0.04)
W = 0.02k ---------------(ii)
Where;
k = 800N/m [as calculated above]
Substitute this value into equation (ii) as follows;
W = 0.02(800)
W = 16J
The work done to stretch the spring by a total extension of 20cm is 16J
Now the additional work required to stretch the spring 10cm further is given by the difference between the total work done and the initial work done as follows;
16 - 4 = 12J
Therefore, the additional work done is 12 J