Ask question

Ask question

All categories

3 years ago

14

An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and

the object eventually comes to a stop. If the spring constant of the spring is 1800 N/m, the initial compression of the spring is 10 cm, the mass of the object is 4 kg, and the coefficient of kinetic friction between the object and the surface is 0.35, how far does the object travel before it comes to a stop?

1 answer:

PolarNik [594]3 years ago

3 0

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

$E_i+K_i+W_f=K_f+U_f$

The work done by the friction force is given by:

$W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]$

so:

$\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m$

You might be interested in

In a fit of rage about distance learning, you throw your computer out the window with an initial velocity of 3 m/s and it breaks

lesya692 [45]

Answer:

9 m

Explanation:

i did the test and got 100%

3 0

3 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

3 years ago

A 60 kg sprinter has a momentum of +600 kg-m/s when he crosses the finish

MakcuM [25]

Answer:

10 ms⁻¹

Explanation:

The amount of momentum that an object has is dependent upon two factors

mass of the moving object
speed of motion

In terms of an equation,

Momentum (P) = Mass(m)×velocity(v)

P = m×v

600 = 60 × v ⇒ v = 10 ms⁻¹

3 0

3 years ago

Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par

kondor19780726 [428]

Answer:

m1/m2 = 0.51

Explanation:

First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:

V = √F/u

This is the equation that describes the relation between speed of a pulse and a force exerted on it.

the value of "u" is:

u = m/L

Where m is the mass of the rod, and L the length.

Now, for the rod 1:

V1 = √F/u1 (1)

rod 2:

V2 = √F/u2 (2)

Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:

1.4V2 = √F/u1 (3)

Replacing (2) in (3):

1.4(√F/u2) = √F/u1 (4)

Now, let's solve the equation 4:

[1.4(√F/u2)]² = F/u1

1.96(F/u2) =F/u1

1.96F = F*u2/u1

1.96 = u2/u1 (5)

Now, replacing the expression of u into (5) we have the following:

1.96 = m2/L / m1/L

1.96 = m2/m1 (6)

But we need m1/m2 so:

1.96m1 = m2

m1/m2 = 1/1.96

m1/m2 = 0.51

5 0

4 years ago

A hose directs a horizontal jet of water, moving with a velocity of 20m/s, on to a vertical wall. The

just olya [345]

Force is defined as the rate of change of momentum.
The initial amount of momentum is $mv$ because water stops when it hit the wall total change of momentum must be $\Delta p=mv$ .
Now let's calculate the force.
$F= \frac{dp}{dt}=\frac{d(mv)}{dt}=\frac{dm}{dt}v$
We need to find $\frac{dm}{dt}$ . This is the amount of water hiting the wall per second.
$\frac{dm}{dt}=\rho Av$
Our final formula would be:
$F=\rho Avv=\rho Av^2$
And now we can calculate the answer:
$F=1000\cdot5\cdot 10^{-4}\cdot(20)^2=200 N$

6 0

3 years ago