Hi there!
Recall the equation for spring potential energy:
![PE = \frac{1}{2}kx^2](https://tex.z-dn.net/?f=PE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
k = Spring constant (N/m)
x = extension of spring from equilibrium (m)
PE = Potential Energy (J)
Plug in the given values:
![PE = \frac{1}{2}(4)(2^2) = \boxed{8 J}](https://tex.z-dn.net/?f=PE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%284%29%282%5E2%29%20%3D%20%5Cboxed%7B8%20J%7D)
Answer:
The Kinetic energy and mass are _directly_ proportional.
Explanation:
We know that Kinetic Energy is basically termed as the capacity of a body to do work.
Kinetic energy is often used to associate with moving objects, therefore, K.E is normally termed as the energy of motion.
The formula of K.E of an object of mass and velocity is defined
K.E = 1/2mv²
From the formula, it is clear that K.E is directly proportional to its mass and also directly proportional to the square of its velocity.
For example,
If A toy plane with a mass of 10 kg is flying at 20 m/s. Its K.E will be:
K.E = 1/2mv²
= 1/2(10)(20)²
= 1/2(10)(400)
= 5(400)
= 2000 J
Now, let suppose, if we double the mass of a toy plane i.e.
m = 20 kg
so
K.E = 1/2mv²
= 1/2(20)(20)²
= 1/2(20)(400)
= 10(400)
= 400 J
Therefore, the K.E is doubled when doubled the mass.
Therefore, the Kinetic energy and mass are _directly_ proportional.
Answer:
37.11231 m/s
Explanation:
u = Initial velocity
v = Final velocity = 0
s = Displacement = 78 m
g = Acceleration due to gravity = 9.81 m/s²
= Coefficient of kinetic friction = 0.9
Acceleration is given by
![a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g](https://tex.z-dn.net/?f=a%3D-%5Cfrac%7Bf%7D%7Bm%7D%5C%5C%5CRightarrow%20a%3D-%5Cfrac%7B%5Cmu%20mg%7D%7Bm%7D%5C%5C%5CRightarrow%20a%3D-%5Cmu%20g)
Equation of motion
![v^2-u^2=2as\\\Rightarrow v^2-u^2=-2\mu gs\\\Rightarrow -u^2=2(-\mu g)s-v^2\\\Rightarrow u=\sqrt{v^2-2(-\mu g)s}\\\Rightarrow u=\sqrt{0^2-2\times (-0.9\times 9.81)\times 78}\\\Rightarrow u=37.11231\ m/s](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20v%5E2-u%5E2%3D-2%5Cmu%20gs%5C%5C%5CRightarrow%20-u%5E2%3D2%28-%5Cmu%20g%29s-v%5E2%5C%5C%5CRightarrow%20u%3D%5Csqrt%7Bv%5E2-2%28-%5Cmu%20g%29s%7D%5C%5C%5CRightarrow%20u%3D%5Csqrt%7B0%5E2-2%5Ctimes%20%28-0.9%5Ctimes%209.81%29%5Ctimes%2078%7D%5C%5C%5CRightarrow%20u%3D37.11231%5C%20m%2Fs)
The initial speed of that car is 37.11231 m/s
Answer:
Tension
upwards
Force
downward
Explanation:
Since the weight of the strut is uniform therefore it can be considered as a uniformly distributed load of 400
over a mass-less beam.
According to the given conditions one end of the strut is attached to a hinge and the other is loaded with a sign of 200 N and supported by a cable in the middle of the span of strut as shown in the schematic.
<u>Now, for the equilibrium condition:</u>
Forces are balanced:
![\sum F_x=0](https://tex.z-dn.net/?f=%5Csum%20F_x%3D0)
![\sum F_y=0](https://tex.z-dn.net/?f=%5Csum%20F_y%3D0)
i.e.
![T+F=200+400](https://tex.z-dn.net/?f=T%2BF%3D200%2B400)
.....................(1)
Moment about any point is balanced:
![\sum M=0](https://tex.z-dn.net/?f=%5Csum%20M%3D0)
<u>Taking moment about the hinge point:</u>
![F\times 0+(T-400)\times \frac{x}{2} = 200\times x](https://tex.z-dn.net/?f=F%5Ctimes%200%2B%28T-400%29%5Ctimes%20%5Cfrac%7Bx%7D%7B2%7D%20%3D%20200%5Ctimes%20x)
upwards
Now put this value in eq. (1)
i.e. negative sign denotes opposite direction to the presumed one.
downward
4.00 is your answer ~Hope this Helps~