Answer:
Explanation:
a ) work done by gravitational force
= mg sinθ ( d + .21)
Potential energy stored in compressed spring
= 1/2 k x²
= .5 x 431 x ( .21 )²
= 9.5
According to conservation of energy
mg sinθ ( d + .21) = 9.5
3.2 x 9.8 x sin 30( d + .21 ) = 9.5
d = 40 cm
b )
As long as mg sin30 is greater than kx ( restoring force ) , there will be acceleration in the block.
mg sin30 = kx
3.2 x 9.8 x .5 = 431 x
x = 3.63 cm
When there is compression of 3.63 cm in the spring , block will have maximum velocity. there after its speed will start decreasing.
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.
I believe it is False, only because the plane is Frictionless. Hope this helps, good luck.
Answer: The normal of the plane must be parallel to the electric field vector.
Explanation:
the normal to the surface is defined as a versor that is perpendicular to the plane.
Now, if the angle between this normal and the line of the field is θ, we have that the flux can be written as:
Φ = E*A*cos(θ)
Where E is the field, A is the area and θ is the angle already defined.
Now, this maximizes when θ = 0.
This means that the normal of the surface must be parallel to the electric field
