A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
1 answer:
Answer: = 16.49N
Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,
= horizontal component of the weight = mgsinФ
= vertical component of weight = mgcosФ
Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.
by using newton's law of motion, we have that
mgsinФ - = ma
where m = mass of object=5kg
a = acceleration= unknown
Ф = angle of inclination = 37°
g = acceleration due to gravity = 9.8
= frictional force = unknown
we need to first get the acceleration before the frictional force which is gotten by using the equation below
where v = final velocity = 2m/s
u = initial velocity = 0m/s (because the object started from rest)
a= unknown
S= distance covered = length of plane = 5m
we slot in a into the equation below to get frictional force
mgsinФ - = ma
3 * 9.8 * sin 37 - = 3* 0.4
17.9633 - = 1.2
= 17.9633 - 1.2
= 16.49N
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Δθ₂h = 690 rev
Explanation:
- Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:
- Now, we need first to find the value of the angular acceleration, that we can get from the following expression:
- Since the machine starts from rest, ω₀ = 0.
- We know the value of ωf₁ (the operating speed) in rev/min.
- Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:
- Replacing by the givens in (2):
- Solving for α:
- Replacing (5) and Δt in (1), we get:
- in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:
- In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:
- First of all, we need to find the value of the angular acceleration during the second period.
- We can use again (2) replacing by the givens:
- ωf =0 (the machine finally comes to an stop)
- ω₀ = ωf₁ = 57.5 rev/sec
- Δt = 32 s
- Solving for α in (9), we get:
- Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:
- In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
Answer:
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