物理計算題 physics calc. 20點!!

2015-01-04 10:25 am
求助~請幫幫忙~~~

Q1
http://postimg.org/image/plsjrvme9/

A block is placed on a slope at an angle of θ = 34.3o as shown in the diagram (the diagram is not to scale).

The height of the inclined plane is h = 1.870 m. The block has a mass of
m = 5.49 kg.
Neglecting friction, what is the speed of the block at the bottom of the slope if it is released from rest at the top of the slope?
v = ___ms-1 at the bottom of the slope

If the coefficient of kinetic friction is μk = 0.111 and the coefficient of static friction is μs = 0.188 what is the speed of the block at the bottom of the slope? Assume it starts from rest at the top of the slope.
__ms-1 (enter "0" (a zero) if it stays at the top)

What is the magnitude of the net force on the block halfway down the slope?
___N

What is the minimum angle, θ, for which the block will start to move down the slope?
___°


Q2

An Olympic pool has dimensions 50.0 m × 25.0 m × 3.13 m (length × width × depth). The density of the pool water is 1015 kgm-3. Atmospheric pressure (1.00 atm) is acting on the surface of the water.

If a child dives down to the bottom of the pool what pressure is exerted on the child?
_ Pa

The child now pulls his air filled blow up toy to the bottom of the pool. The toy is compressible and has a volume of 2.47 L on the surface of the pool. What volume does it have on the bottom of the pool? The air around the pool is at thermal equilibrium with the water in the pool.

_L

Thanks so much!!!!!
更新1:

Thank you so much~~~~~~~ 不過還想補充問一下Q1 的一小題: What is the magnitude of the net force on the block halfway down the slope? ___N 謝謝!!!

回答 (1)

2015-01-04 7:31 pm
✔ 最佳答案
1. Speed = square-root(2g x 1.87) m/s = 6.12 m/s
where g is the acceleration due to gravity, taken to be 10 m/s^2

Frictional force = 5.49g.cos(34.3) x 0.111 N = 5.034 N
Net force = (5.49g.sin(34.3) - 5.034) N = 25.9 N
Acceleration = 25.9/5.49 m/s^2 = 4.718 m/s^2
Speed at bottom of slope = square-root(2 x 4.718 x 1.87/sin(34.3)) m/s
= 5.6 m/s

Max static friction = 5.49g.cos(θ) x 0.188 N
Hence, 5.49g.cos(θ) x 0.188 = 5.49g.sin(θ)
tan(θ) = 0.188
θ = 10.6 degrees

2. Pressure = 3.13 x 1015g Pa = 31770 Pa

New volume = 2.47 x (31770+10^5)/10^5 L = 3.25 L
where 10^5 Pa is the standard atmospheric pressure

2015-01-04 13:19:55 補充:
sorry, a mistake in Q2.
The new volume, using Boyle's Law, is = 2.47 x [10^5/(31770 + 10^5) L = 1.87 L

2015-01-05 13:00:33 補充:
Your suppl question:
The net force has already been found when calculating the acceleration. It is 25.9 N


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