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2 years ago

A spring toy jumps up from the floor and comes back down to the floor. Its initial speed is 12 m/s. What is the

Physics

1 answer:

dybincka [34]2 years ago

8 0

Answer:

$0\; \rm m \cdot s^{-1}$ .

Explanation:

In a projectile motion, the kinetic energy (KE) of the projectile is converted to gravitational potential energy (GPE) and then from GPE back to KE.

In this example, the spring toy is the projectile. Refer to the diagram attached.

The spring toy started with a velocity of $12\; \rm m\cdot s^{-1}$ , meaning that its initial KE is non-zero.
On the way up, the KE of this spring toy gets converted to GPE.
At the top of the trajectory, the GPE of this toy is maximized while its KE is minimized (zero).
As the toy returns to the ground, the GPE of this toy gets converted back to KE.
The GPE of this toy is zero when the toy is on the ground.

In other words, the kinetic energy (KE) of this toy would be $0$ when it is at the top of the trajectory and GPE is maximized.

Since the KE of this toy at the top of the trajectory would be $0$ , the velocity of this toy at that moment would also be $0\!$ .

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3 years ago

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insens350 [35]

The answers is
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3 years ago

A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at

Vesna [10]

Answer:

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0 So B=0

-16 B-32 A+16B=6 So A=-3/16

y=-3/16 cos 4t

Now to find the CF of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

$m^2+8m+16=0$

So $CF =(C_1+tC_2)e^{-2t}$

So the general equation

$Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t$

Given that t=0 Y=0 So

$C_1=\dfrac{3}{16}$

t=0 Y'=0 So

$C_2 =\dfrac{3}{8}$

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

The above equation is the general equation for motion.

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3 years ago

Obtain the formula for the focal length of a lens in terms of object distance (u)

Alexxx [7]

Answer:

m=image distance÷object distance

6 0

2 years ago

In our first example we will consider a very simple application of Newton’s second law. A worker with spikes on his shoes pulls

sweet-ann [11.9K]

Answer:

Acceleration= $0.5m/s^2$

Speed=0.67 m/s

Explanation:

We are given that

Horizontal force=F=20 N

Mass of box=m=40 kg

We know that

Acceleration= $a=\frac{F}{m}$

Using the formula

Acceleration of box= $\frac{20}{40}=0.5m/s^2$

The acceleration of the box= $0.5m/s^2$

Initial velocity=u=0

Force=F=30 N

Distance=s=0.3 m

$a=\frac{30}{40}=\frac{3}{4} ms^{-2}$

$v^2-u^2=2as$

Substitute the values

$v^2-0=2\times \frac{3}{4}\times 0.3=0.45$

$v^2=0.45$

$v=\sqrt{0.45}=0.67m/s$

Hence, the speed of the box after it has been pulled a distance of 0.3 m=0.67 m/s

4 0

3 years ago