spring highschool

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# h1psp1

A 150 g. mass is suspended from a vertical spring and descends a distance of 4.6 cm, after which it hangs at rest. An additional 500 g. mass is then suspended from the first.   What is the total extension of the spring? (neglect the mass of the spring.)

Strategy - the rule "Must have" of Mr.Zhang ®
Find the value of k ;    Fs=kx=mg	→ (1)
And later find x(total stretch length) from ;   kx=(m1+m2)g	→ (2)

Solution

Find k(constant of spring) from equation(1)

k	=	m1g x1
	=	(0.15 kg)(9.8 m·s-2) 0.046 m
	=	32 N·m-1

Find x(total extension of the spring) from equation(2)

x	=	(m1+m2)g k
	=	(0.15 kg+0.50 kg)(9.8 m·s-2) 32 N·m-1
	=	0.20 m(or 20 cm.)

Ans.    The total extension of spring is 20 cm

Brake Booster

Projectile motion

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Physics High School
# h2phyt1 - Projectile motion

A stone is thrown from the edge of a vertical cliff with a velocity of 50 m•s -1

at an angle tan-1

7 24

above the horizontal.    The stone strikes the sea at

a point 240 m from the foot of the cliff.    Find the time for which the stone

is in the air and the height of the cliff.

Strategy - the rule "Must have" of Mr.Zhang ®

1. Convert tan-1

7 24

to be sinθ and cosθ

2. Formulae : Sy = vyt +

1 2

gt2

(vertical) → (1)

S x= vxt (horizontal) → (2)

Horizontal motion (time) = Vertical motion (time)

Solution

Compute the time to find the height

∵   tan-1	7 24
Therefore,tanθ=	7 24	,  sinθ=	7 25	,  cosθ=	24 25

Horizontal motion from equation (2);

S x	=	vxt
S x	=	vxcosθt
240 m	=	50 m•s-1	24 25	t
t	=	5 s

We bring the horizontal time to calculate the vertical
motion to get the height value.

Sy = vyt +

1 2

gt2

→ (1)

Sy = 50 m•s-1sinθ(5) +

1 2

(-9.8 m•s-2)(52)

Gravitational against downward direction of the earth, therefore, we take the minus sign.

-h = 50 m•s-1

7 25

(5) +

1 2

(-9.8 m•s-2)(52)

Displacement direction required is lower the level of the cliff. Hence we take the minus sign.

Ans. 1. The stone is in the air for 5 second.

2. The height of the cliff 52.5 m.

Fuel Injector