(a) The free body of all the forces include, frictional force, weight of the box acting perpendicular and another acting parallel to the plane.
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
<h3>
Free body diagram</h3>
The free body diagram of all the forces on the box is obtained by noting the upward force and downward forces on the box as shown below;
/ W2
Ф → Ff
↓W1
where;
- Ff is the frictional force resisting the down motion of the box
- W1 is the perpendicular component of the box weight = Wcos(33)
- W2 is the parallel component of the box weight = Wsin(33)
(b) When the box is sliding down, the frictional force acts towards the right.
(c) When the box slides up, the direction of the frictional force changes, it acts towards the left.
Learn more about free body diagram of inclined objects here: brainly.com/question/4176810
Answer:
<h2>
d₂ = 3d</h2><h2>
The diameter of the second wire is 3 times that of the initial wire.</h2>
Explanation:
Using the formula for calculating the resistivity of an object to find the diameter.
Resistivity P = RA/L
R is the resistance of the material
A is the cross sectional area
L is the length of the material
Since A = πd²/4
P = R( πd²/4)/L
P = Rπd²/4L ... 1
If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;
P₂ = (R/9)A₂/L₂
P₂ = (R/9)(πd₂²/4)/L₂
P₂ = (Rπd₂²/36)/L₂
P₂ = (Rπd₂²)/36L₂
Since the length and resistivity are the same;
P = P₂ and L =L₂
Equating 1 and 2;
Rπd²/4L = (Rπd₂²)/36L₂
Rπd²/4L = (Rπd₂²)/36L
d² = d₂²/9
d₂² = 9d²
Taking the square root of both sides;
√d₂² = √9d²
d₂ = 3d
Therefore the diameter of the second wire is 3 times that of the initial wire
Answer:
Explanation:
Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:
Where 
are linear kinetic energy, gravitational potential energy and work, respectively.
