Ask question

Ask question

All categories

2 years ago

14

A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m.

When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?

1 answer:

Elza [17]2 years ago

6 0

Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant $k=4.45\times 10^3 N/m$

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

$P_i=\frac{kx^2}{2}$

$P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J$

$P_f=mgh=0.25\times 9.8\times h$

$P_i=P_f$

$14.24 =0.25\times 9.8\times h$

$h=\frac{14.24}{0.25\times 9.8}=5.81 m$

You might be interested in

Give an example of where you see the law of conservation of energy in your life

lesya692 [45]

Answer:

The law of conservation of energy can be seen in these everyday examples of energy transference: Water can produce electricity. Water falls from the sky, converting potential energy to kinetic energy.

The cue ball loses energy because the energy it had has been transferred to the 8 ball, so the cue ball slows down

Explanation:

8 0

2 years ago

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0

2 years ago

A jogger moves from a position x =

vlada-n [284]

Answer:

3 m/s

Explanation:

We'll begin by calculating the change in displacement of the jogger. This can be obtained as follow:

Initial displacement (d₁) = 4 m

Final displacement (d₂) = 16 m

Change in displacement (Δd) =?

Δd = d₂ – d₁

Δd = 16 – 4

Δd = 12 m

Finally, we shall determine the determine the average velocity. This can be obtained as follow:

Change in displacement (Δd) = 12 m

Time (t) = 4 s

Velocity (v) =?

v = Δd / t

v = 12 / 4

v = 3 m/s

Thus, the average velocity of the jogger is 3 m/s

5 0

2 years ago

You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

$F=\frac{kQq}{r^2}$

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

$\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}$

And the force on the charge due to the charge on the north is,

$\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}$

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of $45^\circ$ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

#SPJ4

3 0

1 year ago

Two balls are thrown against a wall. Ball 1 has a much higher speed than ball 2.

Sunny_sXe [5.5K]

Let both the balls have the same mass equals to m.

Let $v_1$ and $v_2$ be the speed of the ball1 and the ball2 respectively, such that

$v_1>v_2\;\cdots(i)$

Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.

The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, $v$ , is $\frac 12 m v^2$

and the potential energy is due to the change in height is $mgh$ [where $g$ is the acceleration due to gravity]

So, the total energy of ball1,

$=\frac 12 m v_1^2 + mgh\;\cdots(ii)$

and the total energy of ball1,

$=\frac 12 m v_2^2 + mgh\;\cdots(iii)$ .

Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)

So, $\frac 12 m v_1^2 >\frac 12 m v_2^2$

Now, from equations (ii) and (iii)

The total energy of ball1 hi higher than the total energy of ball2.

6 0

2 years ago