Complete Question:
A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.
Answer:
Power = 54.07 W
Explanation:
Mass of the block = 10 kg
Angle made with the horizontal, θ = 60°
Distance covered, d = 5 m
Tension in the rope, T = 40 N
Coefficient of kinetic friction, ![\mu = 0.2](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.2)
Let the Normal reaction = N
The weight of the block acting downwards = mg
The vertical resolution of the 40 N force, ![f_{y} = 40sin \theta](https://tex.z-dn.net/?f=f_%7By%7D%20%3D%2040sin%20%5Ctheta)
![\sum f(y) = 0](https://tex.z-dn.net/?f=%5Csum%20f%28y%29%20%3D%200)
![N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N](https://tex.z-dn.net/?f=N%20%2B%2040%20sin%20%5Ctheta%20-%20mg%20%3D%200%5C%5CN%20%3D%20-40sin60%20%2B%2010%2A9.81%20%3D%200%5C%5CN%20%3D%2063.46%20N)
![\sum f(x) = 0](https://tex.z-dn.net/?f=%5Csum%20f%28x%29%20%3D%200)
![40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}](https://tex.z-dn.net/?f=40%20cos%2060%20-%20f_%7Br%7D%20-%20ma%20%3D%200%5C%5C%20f_%7Br%7D%20%3D%20%5Cmu%20N%5C%5C%20f_%7Br%7D%20%3D%200.2%20%2A%2063.46%5C%5C%20f_%7Br%7D%20%3D%2012.69%20N%5C%5C40cos%2060%20-%2012.69-10a%20%3D%200%5C%5C7.31%20%3D%2010a%5C%5Ca%20%3D%200.731%20m%2Fs%5E%7B2%7D)
![v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%20%3D%20u%5E%7B2%7D%20%2B%202as%5C%5Cu%20%3D%200%20m%2Fs%5C%5Cv%5E%7B2%7D%20%20%3D%20%202%20%2A%200.731%20%2A%205%5C%5Cv%5E%7B2%7D%20%20%3D%207.31%5C%5Cv%20%3D%20%5Csqrt%7B7.31%7D%20%5C%5Cv%20%3D%202.704%20m%2Fs)
Power, ![P = Fvcos \theta](https://tex.z-dn.net/?f=P%20%3D%20Fvcos%20%5Ctheta)
![P = 40 *2.704 cos60\\P = 54.074 W](https://tex.z-dn.net/?f=P%20%3D%2040%20%2A2.704%20cos60%5C%5CP%20%3D%2054.074%20W)
D. convergent plate boundary involving an oceanic plate
Red clothes look red because they REFLECT the red light, and absorb light of other colors.
Answer: 313920
Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.
Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.
The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.
The depth below the water surface of each strip is,
di = 8 − yi
and that in turn gives us the pressure on the strip,
Pi =ρgdi = 9810 (8−yi)
The area of each strip is,
Ai = 2√4− (yi) 2Δy
The hydrostatic force on each strip is,
Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy
The total force on the plate is found on the attached image.
Elastic potential energy.
When you stretch a rubber band it has the "potential" to do work, to fly in a given direction. In doing so it changes it's elastic potential energy to kinetic energy.