A box is being pulled to the right. What is the direction of the gravitational force?

An object of mass 2kg is on an incline where there is an applied force of 15N. The

seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is $2.46 m/s^2$

Explanation:

a)

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

The weight of the object, labelled as $mg$ (where m is the mass of the object and g is the acceleration of gravity), acting downward
The applied force, $F_a$ , acting up along the plane
The force of friction, $F_f$ , acting down along the plane

b)

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

$F_a - F_f - mg sin \theta = ma$

where:

$F_a = 15 N$ is the applied force, pushing forward

$F_f = 5 N$ is the frictional force, acting backward

$mg sin \theta$ is the component of the weight parallel to the incline, acting backward, where

m = 2 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=15^{\circ}$ is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)

a is the acceleration

Solving for a, we find:

$a=\frac{F_a - F_f - mg sin \theta}{m}=\frac{15-5-(2)(9.8)(sin 15^{\circ})}{2}=2.46 m/s^2$

Learn more about inclined planes:

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8 0

2 years ago

Example of the center of the gravity<br>

velikii [3]

Answer:

The example of the center of the gravity is the middle of a seesaw

Explanation:

I hope this will help you and plz mark me brainlist

5 0

2 years ago

A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?

iren2701 [21]

Answer:

$K.E=29.403125J$

Explanation:

From the question we are told that

Mass $M=100$

Height $50-20=30m$

Generally the equation for velocity before impact is is is mathematically given by

$v=\sqrt{2gh}$

$v=\sqrt{2*9.8*30}$

$v=24.25$

Generally the equation for Kinetic Energy is is mathematically given by

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

$K.E=\frac{1}{2}*100*(24.25)^2\\$

$K.E=29403.125J$

$K.E=29.403125J$

8 0

2 years ago

ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of

sesenic [268]

Answer:

a) $V_{2/1}=0.8m/s$

b) The second runner will win

c) d = 10.54m

Explanation:

For part (a):

$V_{2/1} = V_{2} - V_{1} = 0.8m/s$

For part (b) we will calculate the amount of time that takes both runners to cross the finish line:

$t_{1} = \frac{X_{1}}{V_{1}}=\frac{250}{3.45}=72.46s$

$t_{2} = \frac{X_{2}}{V_{2}}=\frac{250+45}{4.25}=69.41s$

Since it takes less time to the second runner to cross the finish line, we can say the she won the race.

For part (c), we know how much time it takes the second runner to win, so we just need the position of the first runner in that moment:

X1 = V1*t2 = 239.46m Since the finish line was 250m away:

d = 250m - 239.46m = 10.54m

6 0

2 years ago

Read 2 more answers

The sparse area surrounding a spiral galaxy is called a

djyliett [7]

<span>The sparse area surrounding a spiral galaxy is called a

Halo

</span>

3 0

3 years ago

Read 2 more answers