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Ugo [173]
2 years ago
13

A 0.650 kg block is attached to a spring with spring constant 18.0 N/m . While the block is sitting at rest, a student hits it w

ith a hammer and almost instantaneously gives it a speed of 47.0 cm/s . What are
The amplitude of the subsequent oscillations? answer is in cm

The block's speed at the point where x= 0.350 A? answer is in cm/s
Physics
1 answer:
garik1379 [7]2 years ago
3 0

Answer:

amplitude is 8.92 cm

speed of block is 44.11 cm/s

Explanation:

given data

mass = 0.650 kg

spring constant = 18 N/m

speed = 47 cm/s = 0.44 m/s

speed at x point = 0.350 A

to find out

amplitude of subsequent oscillation

solution

we know here conservation of energy

maximum kinetic energy = maximum potential energy

\frac{1}{2} m v^{2} = \frac{1}{2} k A^{2}

here we know k is spring constant and m is mass and A is amplitude and v is velocity

so solve it we get

A = \sqrt{\frac{mv^{2} }{k} }

put here all these value

A =  \sqrt{\frac{0.650(0.47)^{2} }{18} }

A = 0.08931 m

so amplitude is 8.92 cm

and

by conservation of energy

initial energy = final energy

\frac{1}{2} m V^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} m Vm^{2} + 0

solve we gey V

V = \sqrt{Vm^{2} - \frac{k }{m} x^{2} }    

put here value Vm = 0.47 , and x = 0.350

V = \sqrt{0.47^{2} - \frac{18 }{0.650} (0.350*0.088 m)^{2} }

V = 0.4411 m/s

so speed of block is 44.11 cm/s

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A spring is used as part of a lift system and follows Hooke's law. If the spring is
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Answer:

1.05m or 105cm

Explanation:

Using the hooke's law equation as follows;

F = –k.x

Where;

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According to the information given in this question;

Displacement (x) = 85cm = 85/100 = 0.85m

Force = 12500N

Using F = kx, we find the proportionality constant

k = F/x

K = 12500/0.85

K = 14705.8N/m.

Also, since K = 14705.8N/m, the displacement (x), when the force increases to 15500N is;

F = kx

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x = 15500/14705.8

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What is the requirement for a core to be used in an electromagnet?
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D. It must be able to be magnetized

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3 years ago
A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.
Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

PE=KE\\mgh = \frac{1}{2}mv^2

where

m = 0.160 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

s=(\frac{u+v}{2})t

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s

C)

The force exerted by the floor on the ball can be found using the equation:

F=\frac{\Delta p}{t}

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\Delta p is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

\Delta p = m(v-u)

So the equation can be rewritten as

F=\frac{m(v-u)}{t}

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

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