A modified Atwood machine is at rest. The hanging block has a mass of 3kg. The black on wheels has unknown mass M1. When release
d the block m2 falls at 5.88m/s^2. If frictional forces are considered so small they are negligible, the block M1 must have a mass of _______ kg.
1 answer:
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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:
Answer:
1.35m
Explanation:
At the highest point of the jump, the vertical speed of the skier should be 0. So the 13m/s speed is horizontal, this speed stays the same from the jumping point to the highest point. The 14m/s speed at jumping point is the combination of both vertical and horizontal speeds.
The vertical speed at the jumping point can be computed:
When the skier jumps to the its potential energy is converted to kinetic energy:
where m is the skier mass and h is the vertical distance traveled, is the vertical velocity at jumping point, and h is the highest point.
Let g = 10m/s2
We can divide both sides of the equation by m: