A 2.0 kgkg wood block is launched up a wooden ramp that is inclined at a 25 ∘∘ angle. The block's initial speed is 11 m/sm/s . The coefficient of kinetic friction of wood on wood is μk=0.200μk=0.200. Part A What vertical height does the block reach above its starting point?
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Ответ:
The B part of the question is missing and it says;
What speed does it have when it slides back down to its starting point?
A) vertical height the block reaches above its starting point = 3.568m
B) Speed of block when it slides back down to its starting point = 6.323 m/s
Explanation:
A) Since inclined at an angle of 30°,thus;
For kinetic friction, h = mg cosθ
Thu kinetic friction force(Fk) is ;
Fk = μmg cosθ
Let's choose the x-direction to be parallel to the ramp surface.
Resolving all the forces on the block in the x-direction;
ΣFx; μmg cosθ = -ma - mg sinθ
Divide each term by m to get;
μg cosθ = -a - g sinθ
Making a the subject of the formula,
a = -μg cosθ - g sinθ
a = - g(μ cosθ + sinθ)
So a = - 9.81[0.2(cos25°) + sin 25°]
a = - 9.81[ 0.1813 + 0.4226]
a = -5.924 m/s²
Using equations of motion,
v² = u² + 2as
Now v = 0m/s while u= 10m/s and s=Δx since ramp is chosen to be parallel to x-direction
Thus 0² = 10² + 2(-5.924)(Δx)
So making (Δx) the subject ;
11.85(Δx) = 100
So, Δx = 100/11.85 = 8.44m
Thus; vertical height which is the vertical component displacement = (Δx) x sinθ = 8.44Sin(25) = 8.44 x 0.4226 = 3.568m
B) when the block is sliding down, the sign of the friction changes.
Thus;
ΣFx; - μmg cosθ = -ma - mg sinθ
So, a = μg cosθ - g sinθ
a = g(μ cosθ - sinθ)
So a = 9.81[0.2(cos25°) - sin 25°]
a = 9.81[ 0.1813 - 0.4226] = - 2.37 m/s²
Again, using equations of motion,
v² = u² + 2as
So in thus case u =0 and since it is moving down the plane, the displacement will be negative, thus
(Δx) = - 8.44m
So, v² = 2(-2.37)(-8.44) = 40
Thus, v = √40 = 6.323 m/s
Ответ:
The answer would be A or D. The most appropriate answer would be D. Hope I helped :)