Let us consider body moves a distance S due to the force F.
Hence the work by the body W = FS
If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -
where
is the component of the force along the direction of displacement.


As per the question the power P is given as -




Hence alternative definition of power P = F.V
I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.
a 10 kg block reaches a point with a velocity of 15 m per second and slides down a rough track my the coefficient of the kinetic energy between the two surface ab and the block iis0.52
Hello!
Most ocean waves obtain their energy and motion from the wind.
Ocean waves are surface waves that move across the surface of the ocean. When wind touches the surface of the water, there is friction in the contact zone. This friction causes a drag effect, that makes wrinkles on the surface of the water. As the wrinkles get bigger, they transform into full-blown waves, and the taller the wave, the more energy it can extract from the wind, making them even bigger and to move longer distances.
Have a nice day!
Answer:
95.9°
Explanation:
The diagram illustrating the action of the two forces on the object is given in the attached photo.
Using sine rule a/SineA = b/SineB, we can obtain the value of B° as shown in the attached photo as follow:
a/SineA = b/SineB,
83/Sine52 = 56/SineB
Cross multiply to express in linear form
83 x SineB = 56 x Sine52
Divide both side by 83
SineB = (56 x Sine52)/83
SineB = 0.5317
B = Sine^-1(0.5317)
B = 32.1°
Now, we can obtain the angle θ, between the two forces as shown in the attached photo as follow:
52° + B° + θ = 180° ( sum of angles in a triangle)
52° + 32.1° + θ = 180°
Collect like terms
θ = 180° - 52° - 32.1°
θ = 95.9°
Therefore, the angle between the two forces is 95.9°