Answer:
I will say that the the potential energy will be at its maximum.
Explanation:
potential energy deals with gravity and gravity deals with height, so when a object is in its maximum height it will have the maximum potential energy.
Answer:
23.52092 J
Explanation:
m = Mass of block = 6.79 kg
s = Sliding distance = 2.82 m
= Angle of slide = 20.7°
= Coefficient of kinetic friction = 0.425
g = Acceleration due to gravity = 9.8 m/s²
Work done by the force of gravity is given by
The work done by the force of gravity is 23.52092 J
Don’t trust those links they usually pull up your IP
Answer:
F₁ = 4,120.2 N
F₂ = 3,924N
Explanation:
1) Balance of angular momentum around the end where F₁ is applied.
F₂ × 0.5m - F₁ × 0 = mass × g × 1m
⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m
F₂ = 196.2 Nm / 0.5m = 3,924 N
2) Balance of forces
F₁ - F₂ = mg
F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N
Answer:
Explanation:
Given that
The mass of the body is 0.04kg
M=0.04kg
The radius of the paths is 0.6m
r=0.6m
The normal force exerted at A is 3.9N
Fa=3.9N
The normal force exerted at B is 0.69N
Fb=0.69N
Then work done by friction from point A to B will be the change in K.E
W=∆K.E+P.E
So we need to know the velocity at both point A and B
Then since the centripetal force is given as
Ft=mv²/r
Then,
For point A
Fa=mv²/r
3.9=0.04v²/0.6
3.9=0.0667v²
v²=3.9/0.0667
v²=58.5
v=√58.5
v=7.65m/s
Va=7.65m/s
Now at point B
Fb=mv²/r
0.69=0.04v²/0.6
0.69=0.0667v²
v²=0.69/0.0667
v²=10.35
v=√10.35
v=3.22m/s
Vb=3.22m/s
Then, the work done is
W=∆K.E+P.E
P.E is given as mgh
The height will be 2R =1.2m
P.E=mgh
P.E=0.04×9.81×1.2
P.E=0.471J
Final kinetic energy at B minus initial kinetic energy at A
W=K.Eb-K.Ea
K.E is given as 1/2mv²
W=1/2m(Vb²-Va²) +P.E
W=0.5×0.04(3.22²-7.65²) +0.471
W=0.5×0.04×(-48.1541) +0.471
W=-0.96+0.471
W=-0.49J
work was done on the block by friction during the motion of the block from point A to point B is 0.49J.
Friction opposes motions and that is why the work done is negative
