Part 2 "Tug-of-War"

Find the kinetic energy of a 0.1-kilogram toy truck moving at the speed of 1.1 meters per second.

ZanzabumX [31]

Answer:

0.061 J

Explanation:

The kinetic energy of an object is given by

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

where

m is the mass of the object

v is its speed

For the toy truck in the problem, we have

m = 0.1 kg is its mass

v = 1.1 m/s is its speed

Putting the numbers into the equation, we find

$K=\frac{1}{2}(0.1 kg)(1.1 m/s)^2=0.061 J$

7 0

3 years ago

A 260-kg glider is being pulled by a 1,940-kg jet along a horizontal runway with an acceleration of a= 2.20 m/s^2 to the right.

lisov135 [29]

Answer:

a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

Explanation:

a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:

Jet

$\Sigma F = F - T = m_{J}\cdot a$ (1)

Glider

$\Sigma F = T = m_{G}\cdot a$ (2)

Where:

$F$ - Thrust of jet engines, measured in newtons.

$T$ - Tension in the cable connecting the jet and glider, measured in newtons.

$m_{G}$ , $m_{J}$ - Masses of the glider and the jet, measured in kilograms.

$a$ - Acceleration of the glider-jet system, measured in meters per square second.

If we know that $m_{G} = 260\,kg$ , $m_{J} = 1,940\,kg$ and $a = 2.20\,\frac{m}{s^{2}}$ , then the solution of this system of equations:

By (2):

$T = (260\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)$

$T = 572\,N$

By (1):

$F = T+m_{J}\cdot a$

$F = 572\,N+(1,940\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)$

$F = 4840\,N$

The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

7 0

3 years ago

When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequenc

Arlecino [84]

Answer:

$m_2/m_1=9.745$

The ratio m_2/m_1 of the masses is 9.745

Explanation:

Formula for frequency when mass m_1 is hung on the spring:

$f_1=\frac{1}{2\pi}\sqrt{\frac{k}{m_1}}$

where:

k is the spring constant

Formula for frequency when mass m_2 is hung on the spring along with m_1:

$f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}$

where:

k is the spring constant.

In order to find ratio m_2/m_1, Divide the above equations:

$\frac{f_1}{f_2} =\frac{ \frac{1}{2\pi}\sqrt{\frac{k}{m_1}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}}$

On Solving the above equation:

$\frac{f_1}{f_2} =\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_1+m_2}}}\\(\frac{f_1}{f_2})^{2} =\frac{m_1+m_2}{m_1} \\(\frac{11.8}{3.60})^2= \frac{m_1+m_2}{m_1} \\10.745=\frac{m_1+m_2}{m_1}\\10.745m_1=m_1+m_2\\m_2=10.745m_1-1m_1\\m_2/m_1=9.745$