All categories

3 years ago

A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless

pulley, as shown in figure (a). The coefficient of kinetic friction between the block and the surface is 0.300.
(a) Find the acceleration of the two objects and the tension in the string.
(b) Check the answer for the acceleration by using the system approach. (Use the following as necessary: m1, m2, μk, and g.)
(c) What if an additional mass is attached to the ball? How large must this mass be to increase the downward acceleration by 60%?

For (a) I calculated acceleration to be 5.10 m/s^2 and tension to be 36.2N.
For (b) I the equation is a = (m2g-ukm1g)/(m1+m2)
It is (c) that I cannot figure out. Can anybody please help me?

Physics

1 answer:

Allisa [31]3 years ago

4 0

$1.6a = \frac{g(m_2 + m_3 - \mu km_1)}{m_1 + m_2 + m_3} \\ \\ 1.6a(m_1 + m_2 + m_3) = g(m_2 + m_3 - \mu km_1) \\ \\ (1.6a + \mu kg)m_1 + (1.6a - g)m_2 = (g - 1.6a)m_3 \\ \\ m_3 = \frac{1.6a +\mu kg}{g - 1.6a} m_1 - m_2 \\ \\ m_3 = 22.57 kg$

You might be interested in

An object at rest starts accelerating.

Shalnov [3]

Answer:

We are given:

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

Solving for 'a'

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

6 0

3 years ago

A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.

dedylja [7]

Answer:

F = 768 N

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, $\dfrac{dm}{dt}=240\ kg/min = 4\ kg/s$

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

$F=\dfrac{dp}{dt}$

$F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$

Since, the speed is constant,

$F=m\dfrac{dv}{dt}$

$F=v\dfrac{dm}{dt}$

$F=3.2\times 240$

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

5 0

3 years ago

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

Ayuda por favor (archivo adjunto) con un ejercicio de expresión sobre periodo de oscilación de esta figura:

vodka [1.7K]

Answer:

nolo se

Explanation:

no lo se

8 0

3 years ago

How do i solve? a = bc (b)

stira [4]

I don’t even know I’m so dumb.

7 0

3 years ago