Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
Answer:
The kinetic energy of the arrow is equaled to the potential energy of the stretching of the bow, which in this case is 50 J.
Explanation:
Potential energy converts to kinetic as soon as it begins to move.
Answer:
When the ocean floors move it makes a wave
Explanation:
Answer:0.12 amperes
Explanation:
voltage=17.6v
Resistance=165.5 ohms
Current=voltage ➗ resistance
Current=17.6 ➗ 165.5
Current=0.12 amperes
Answer:
W = 320.30 J
Explanation:
To calculate the net work done over the block you take into account all implied forces:
(1)
The gravitational force and friction force are against the applied force F.
θ = 26°
F: applied force = 340N
Fg: gravitational force = Mg = (40.0kg)(9.8m/s^2) = 392N
Ff: friction force = 
Next, you replace to obtain the net force:

Finally, the net work, for 4 m, is:
