Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
Here is the highly detailed, arcane, complex, technical form of Ohm's Law that is needed in order to answer this question ===> I = V / R .
Current = (voltage) / (resistance)
Current = (1.5 V) / (10 Ω)
<em>Current = 0.15 Ampere</em>
Answer:
10m/s
Explanation:
Using the law of conservation of momentums
M1u1+m2u2 = (m1+m2)v
Substitute.
4000(10)+1500(10) = (4000+1500)v
40,000+15,000 = 5,500v
55000 = 5500v
v = 55000/5500
v= 10m/s
Hence the velocity of the truck after Collision is 10m/s
