All categories

3 years ago

A student pulls on a 20 kg box with a force of 50 N at an angle 45 degrees relative to the horizontal. The box increases

Physics

1 answer:

telo118 [61]3 years ago

8 0

The value of the friction force is 5.4 N

Explanation:

In order to solve the problem, we have to write down the equation of the forces along the horizontal direction.

We have only two forces acting along the horizontal direction:

The horizontal component of the pull of the student, $F cos \theta$ , forward
The friction force, $F_f$ , backward

The resultant force must be equal to the product between mass and acceleration (Newton's second law), so we can write:

$F cos \theta - F_f = ma$

where:

F = 50 N is the force of pull of the student

$\theta=45^{\circ}$ is the angle relative to the horizontal

m = 20 kg is the mass of the box

$a=1.5 m/s^2$ is the acceleration

Solving the equation for $F_f$ , we find the value of the friction force:

$F_f = Fcos \theta -ma=(50)(cos 45^{\circ})-(20)(1.5)=5.4 N$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

You might be interested in

Which term describes a gap in the geologic record that occurs when sedimentary rocks cover an eroded surface?

Lemur [1.5K]

The term that describes a gap in the geologic record that occurs when sedimentary rocks cover an eroded surface is called unconformity.

Glad to help ya!! ;)

3 0

3 years ago

Read 2 more answers

And 8 kg bowling ball is rolling along the frictionless alley

VLD [36.1K]

It will stop eventually

8 0

3 years ago

What is the only bone of the skull that moves?

Darina [25.2K]

Answer:

Mandible.

Explanation:

The mandible in the bone that allows you to open and close your mouth, otherwise known ad the jawbone.

3 0

3 years ago

Read 2 more answers

A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.

alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

$F_g = F_n + F_a$

given that

$F_g = mg$

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

$F_n = 150 N$

now from above equation

$30*10 = 150 + F_a$

$F_a = 300 - 150 = 150 N$

So force applied by the person must be 150 N

7 0

3 years ago

Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.

Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = $\frac{mv}{qB}$

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r = $\frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T}$ = 0.0328 m, or 3.28 cm.

8 0

3 years ago