All categories

3 years ago

What is the acceleration of a block on a ramp inclined 35o to the horizontal if µk = 0.4?

1 answer:

lina2011 [118]3 years ago

7 0

We can solve the problem by applying Newton's second law, which states that the resultant of the forces acting on an object is equal to the product between its mass and its acceleration:
$\sum F = ma$

We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):

$mg \sin \theta - \mu_K N = ma$ (parallel direction) (1)
$mg \cos \theta - N =0$ (perpendicular direction) (2)
where
$\theta=35^{\circ}$ is the angle of the inclined plane, N is the normal reaction of the plane, $\mu_K N$ is the frictional force, with $\mu_K=0.4$ being the coefficient of friction.

From eq.(2), we find
$N=mg \cos \theta$
and if we substitute into eq.(1), we can find the acceleration of the block:
$mg \sin \theta - \mu_k mg \cos \theta = ma$
from which
$a=g(\sin \theta - \mu_K \cos \theta)=(9.81 m/s^2)(\sin 35^{\circ} - 0.4 \cos 35^{\circ})=2.41 m/s^2$

You might be interested in

Assume that a gravitational anomaly in the solar system has shifted a field of asteroids into Earth’s orbit, and the field is no

Mandarinka [93]

Answer:

An asteroid is a minor planet of the inner Solar System. Historically, these terms have been applied to any astronomical object orbiting the Sun.

7 0

2 years ago

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

75 kmh⁻¹

Explanation:

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago

An elevator manufacturing company is stress-testing a new elevator in an airless test shaft. The elevator is traveling at an unk

devlian [24]

Answer:

3.192 m/s

Explanation:

t = Time taken = 0.900 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 1.1 meters

a = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1.1-\frac{1}{2}\times 9.81\times 0.9^2}{0.9}\\\Rightarrow u=-3.192\ m/s$

Velocity of the elevator when it snapped is 3.192 m/s

4 0

3 years ago

Plz help with these 4

Airida [17]

Answer:

1) The human skeleton performs six major functions: support, movement, protection, production of blood cells, storage of minerals, and endocrine regulation. protection of internal organs

2) Joints are where two bones meet. They make the skeleton flexible — without them, movement would be impossible. Joints allow our bodies to move in many ways.

3)A joint is a point where two or more bones meet. There are three main types of joints; Fibrous (immovable), Cartilaginous (partially moveable) and the Synovial (freely moveable) joint

4)A ligament is a fibrous connective tissue which attaches bone to bone, and usually serves to hold structures together and keep them stable.

Explanation:

go-gle your welcome ;)

5 0

3 years ago

Read 2 more answers

3. With a<br> you can measure just about anything.

Degger [83]

MULTIMETER combines the functions of ammeter, voltmeter and ohmmeter as a minimum.

5 0

3 years ago