How much work is done if a force of 20 N moves an object a distance of 6 m?
1 answer:
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Answer:
<em>The total length of the spring would be 0.65 m</em>
Explanation:
The Concept
Hooke's law evaluates the increment of spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;
F = k x ..................1
where F is the force applied to the spring
k is the spring constant
x is the spring stretch or extension
Step by Step Calculations
We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;
x = F/k
but F = m x g
so, x = (m x g)/k
given that, the mass of the person m =150 kg
g is the acceleration due to gravity = 9.81 m/
k is the spring constant = 10000 N/m
then x = (9.81 m/ x 150 kg)/10000 N/m
x = 0.14715 m
the extension experienced by the spring after the compression is 0.14715 m
The total length of the spring would be;
L = 0.14715 m + 0.5 m = 0.64715
L ≈ 0.65 m
Therefore the total length of the spring would be 0.65 m
Answer:
A)
B)
Explanation:
Given that
Force = F
Increase in Kinetic energy =
we know that
Work done by all the forces =change in the kinetic energy
a)
Lets distance = d
We know work done by force F
W= F .d
F.d=ΔKE
b)
If the force become twice
F' = 2 F
F'.d=ΔKE'
2 F .d = ΔKE' ( F.d =Δ KE)
2ΔKE = ΔKE'
Therefore the final kinetic energy will become the twice if the force become twice.