Start Point: an unlit match. End Point: a lit match.
1 answer:
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Answer:
<h3>0.69</h3>Explanation:
Using the Newtons law of motion;
Fm is the moving force = 400N
Ff is the frictional force = μR
μ is the coefficient of kinetic friction
R is the reaction = mg
m is the mass
a is the acceleration
The equation becomes;
Hence the coefficient of kinetic friction between the box and floor is 0.69
Explanation:
<em>(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively. Each force must be represented by a distinct arrow starting on and pointing away from the dot.</em>
At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.
At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.
<em>(b) i. Derive an expression for the speed of the sphere at point A.</em>
Energy is conserved:
PE = PE + KE + RE
mgH = mgR + ½mv² + ½Iω²
mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²
mgH = mgR + ½mv² + ⅕mv²
gH = gR + ⁷/₁₀ v²
v² = 10g(H−R)/7
v = √(10g(H−R)/7)
<em>ii. Derive an expression for the normal force the track exerts on the sphere at point A.</em>
Sum of forces in the radial (-x) direction:
∑F = ma
N = mv²/R
N = m (10g(H−R)/7) / R
N = 10mg(H−R)/(7R)
<em>(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.</em>
RE / KE
= (½Iω²) / (½mv²)
= ½(⅖mr²)(v/r)² / (½mv²)
= (⅕mv²) / (½mv²)
= ⅕ / ½
= ⅖
<em>(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin. The sphere is replaced with a hoop of the same mass and radius. Will the value of Hmin increase, decrease, or stay the same? Justify your answer.</em>
When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0. Sum of forces in the radial (-y) direction:
∑F = ma
mg = mv²/R
gR = v²
Applying conservation of energy:
PE = PE + KE + RE
mgH = mg(2R) + ½mv² + ½Iω²
mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²
mgH = 2mgR + ½mv² + ½kmv²
gH = 2gR + ½v² + ½kv²
gH = 2gR + ½v² (1 + k)
Substituting for v²:
gH = 2gR + ½(gR) (1 + k)
H = 2R + ½R (1 + k)
H = ½R (4 + 1 + k)
H = ½R (5 + k)
For a sphere, k = 2/5. For a hoop, k = 1. As k increases, H increases.
<em>(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above. The track makes an angle of θ above the horizontal at point C. Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate. Calculate the maximum height above the bottom of the loop that the sphere will reach.</em>
C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.
v₀ = √(10g(H−R)/7)
The vertical component of this initial velocity is v₀ sin θ. At the maximum height, the vertical velocity is 0. During this time, the sphere is in free fall. The maximum height reached is therefore:
v² = v₀² + 2aΔx
0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)
0 = 10g(H−R)/7 sin²θ − 2g(h − R)
2g(h − R) = 10g(H−R)/7 sin²θ
h − R = 5(H−R)/7 sin²θ
h = R + ⁵/₇(H−R)sin²θ
