If 5J of work are done on a spring, compressing it by 12cm, what is the spring constant?

You and a friend work in buildings five equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely

Pavel [41]

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

$s_1 = v_1t = 1.6t$

Similarly the distance (m) from your friend building to the restaurant:

$s_2 = v_2t = 1.2t$

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

$s_1 + s_2 = 5b$

$1.6t + 1.2t = 5b$

$2.8t = 5b$

$t = 5b/2.8 = 25b/14$

$s_2 = 1.2t = 1.2(25b/14) = 2.143b$

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

7 0

3 years ago

Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

max2010maxim [7]

Answer:

The spring was compressed the following amount:

$\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount $\Delta x$ ), and the total final energy is the addition of the kinetic energies of both masses:

$E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2$

$E_i=E_f\\$

$\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

8 0

3 years ago

Suppose you have a 136-kg wooden crate resting on a wood floor. The coefficient of static friction is 0.4 and coefficient of kin

stealth61 [152]

Answer:

The the maximum force acting on the crate is 533.12 newtons.

Explanation:

It is given that,

Mass of the wooden crate, m = 136 kg

The coefficient of static friction, $\mu_s=0.4$

The coefficient of kinetic friction, $\mu_k=0.2$

We need to find the maximum force exerted horizontally on the crate without moving it. As the crate is not moving than the coefficient of static friction will act and the force is given by :

$F=\mu_s mg$

$F=0.4\times 136\ kg\times 9.8\ m/s^2$

F = 533.12 N

So, the maximum force acting on the crate is 533.12 newtons. Hence, this is the required solution.

4 0

3 years ago

Which type of motor is shown below?

Viefleur [7K]

Answer:

AC motor........A

Explanation:

Plato, an AC motor is much simpler than a DC motor. As alternating current passes through the coil, it becomes magnetized. Since the current is alternating, the north and south ends keep switching.

5 0

3 years ago

Read 2 more answers

Which best summarizes a concept related to the work-energy theorem?

Elenna [48]

Answer: When work is positive, the environment does work on an object

Explanation:

5 0

3 years ago