Answer:
W = 30 J
Explanation:
given,
Work done = 10 J
Stretch of spring, x = 0.1 m
We know,
dW = F .dx
we know, F = k x
![W = k[\dfrac{x^2}{2}]_0^{0.1}](https://tex.z-dn.net/?f=W%20%3D%20k%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_0%5E%7B0.1%7D)
k = 2000
now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.
![W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx](https://tex.z-dn.net/?f=W%20%3D%202000%20%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_%7B0.1%7D%5E%7B0.2%7D%20dx)
W = 1000 x 0.03
W = 30 J
Hence, work done is equal to 30 J.
Answer:
a) V1=11.05m/s V2=92.07m/s V3=17.24m/s
b) KE = 16238.26J
Explanation:
For tangential speeds:
For the kinetic energy, it can be calculated as:
Where:
So,
KE=16238.26J
Answer:
5 feet
Explanation:
First lets imagine that the values are in meters and kg, to make it easier for me.
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The force created by the teacher is 120 × 5 = 600N.
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The force created by the 50 kg student is 50 × 8 = 400N
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The force that needs to be created by the 40 kg student is 600N - 400N = 200N.
To create that force that student needs to sit 200 ÷ 40 = 5m.
So the 40 lbs student is sitting 5 feet from the fulcrum.
