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son4ous [18]
3 years ago
8

You attach a 1.70 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by

0.200 m and release it from rest. Assume the block slides on a horizontal surface with negligible friction. The block reaches a speed of zero again 0.200 s after release (for the first time after release). What is the maximum speed of the block (in m/s)
Physics
1 answer:
Usimov [2.4K]3 years ago
8 0

Answer:

maximum speed of the block is 3.14 m/s

Explanation:

given data

mass = 1.70 kg

stretch in the spring = 0.2 m

time take by block to come to zero  t = 0.2 s

solution

we know that Time period of oscillation (T) that is express as

T = 2t    ......................1

put here value

T = 2 (0.2)

T = 0.4 s

so here time period is express as

T = 2\pi \sqrt{\frac{m}{k}}     ................2

here k is spring constant of the spring so  put here value

0.4  =  2(\pi ) \sqrt{\frac{1.70}{k}}  

here k will be

k = 419.02 N/m

so we use here conservation of energy that is

Maximum kinetic energy = Maximum spring potential energy   ............3

(0.5) m v² = (0.5) k x²

here v is maximum speed block

so put here value and we get

(1.70) v² = (419.02) (0.2)²

v = 3.14 m/s

so maximum speed of the block is 3.14 m/s

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A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho
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Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

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Considering gravity, g = 9.8 ms^-^2

And

Using, sin(18) =0.30 and cos(18)=0.95

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

a.

When the man pushes it parallel to the incline.

Balancing the forces as  \sum F=0 .

⇒ F+mgsin(\theta) =0

⇒ F=-mgsin(\theta)

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b.

When the force is parallel to the floor.

⇒ Fcos(\theta)=mgsin(\theta)

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⇒ F=\frac{190\times 9.8\times sin(18)}{cos(18)}

⇒ F=605 N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

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