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worty [1.4K]
3 years ago
9

A horizontal mass-spring system is in SHM. The mass is 0.25kg, the spring constant is 15N/m, and the amplitude is 20cm. a)What i

s the maximum speed of the mass and where does it occur? b) What is its speed at the 1/2 amplitude position of its motion?
Physics
1 answer:
Alex_Xolod [135]3 years ago
3 0

Answer:

maximum speed = 1.55 m/s

speed is 0.775 m/s

Explanation:

given data

mass m = 0.25 kg

spring constant k = 15 N/m

amplitude A = 20 cm = 0.2 m

to find out

maximum speed and speed at the 1/2 amplitude

solution

we know formula for maximum speed that is

maximum speed = amplitude ×√(k/mass)

put all value

maximum speed = 0.2 ×√(15/0.25)

maximum speed = 1.55 m/s

and

at the 1/2 amplitude

we apply here conservation of energy that is

1/2 × m×v² = 1/2 × k×x²

here we know x = A/2 that is 20 /2 = 10 cm = 0.1 m

so

0.25 v² = 15 × 0.1²

v² = 0.600625

v = 0.775

so speed is 0.775 m/s

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4 years ago
Read 2 more answers
Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim
Nonamiya [84]

Answer:

The distance covered by puck A before collision is  z = 8.56 \ m

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  v_A =  3.90 \ m/s

        The speed of puck B is  v_B  =  4.30 \ m/s

The distance covered by puck A is mathematically represented as

     z =  v_A * t

  =>  t  =  \frac{z}{v_A}

 The distance covered by puck B  is  mathematically represented as

      18 - z =  v_B  * t

=>   t  = \frac{18 - z}{v_B}

Since the time take before collision is the same

        \frac{18 - z}{V_B}  =  \frac{z}{v_A}

substituting values

          \frac{18 -z }{4.3}  = \frac{z}{3.90}

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8 0
3 years ago
A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l
mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be \who_w = 1000 kg/m^3

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

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6 0
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