Plz help me with question 3
2 answers:
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Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
![W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7BL%7D%20Fdx%20%3D%20%5Cint_%7B0%7D%5E%7BL%7Dax%20%5C%2C%20dx%20%3D%20%5Cfrac%7Ba%7D%7B2%7D%20%20%5Bx%5E%7B2%7D%20%5D_%7B0%7D%5E%7BL%7D%20%3D%20%20%5Cfrac%7BaL%5E%7B2%7D%7D%7B2%7D%20)
Answer:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
Answer:
Recall that
The car starts from rest, so , and we get
Explanation:
Given that,
Mass of the block, m = 12.2 kg
Initial velocity of the block, u = 6.65 m/s
The coefficient of kinetic friction,
(a)The force of kinetic friction is given by :
mg is the normal force
So,
(b) Net force acting on the block in the horizontal direction,
f = ma
a is the acceleration of the block
(c) Let d is the distance covered by the block before coming to the rest. Using third equation of motion as follows :
Hence, this is the required solution.