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Ganezh [65]
3 years ago
9

Jonny is pushing a 50-kg large package along the dorm room floor with 150 N force. Both the direction of push and the floor are

horizontal. While he is pushing the package, it moves in a straight line with constant 0.6 m/s speed. Please ignore air-resistance and make sure to justify your answers to the questions below.
a) (10 points) Write down as much information as you can about the main horizontal forces exerted on the package (names, directions, and if enough information is given, numerical values).
b) (10 points) Write down as much information as you can about the main vertical forces exerted on the package (names, directions, and whenever possible, if enough information is given, numerical values).
c) (5 points) How far did the package move during 3 seconds of pushing? Explain
d) (5 points) What was the package’s acceleration during that time? Explain
2
Physics
1 answer:
eduard3 years ago
6 0

a. The push is 150 N and acts to the right while the frictional force is 150 N and acts to the left.

b. The weight is 490 N and acts downwards while the normal force is 490 N and acts upwards.

c. The package moves 1.8 m

d. The package's acceleration is 0 m/s²

<h3>a. The horizontal forces</h3>

The push is 150 N and acts to the right while the frictional force is 150 N and acts to the left.a

Since the direction of push and the floor is horizontal, and first horizontal force acting on the package is the push and its magnitude is 150 N.

Also, a frictional force also acts to oppose the motion of the package.

Since the packge moves at a constant velocity of 0.6 m/s, its acceleration is zero and thus the net force on the package is zero.

Let

  • F = push force and
  • f = frictional force

So, F - f = 0

F = f

= 150 N

So, the frictional force is 150 N and opposite to the push.

So, the push is 150 N and acts to the right while the frictional force is 150 N and acts to the left.

<h3>b Vertical forces on package</h3>

The weight is 490 N and acts downwards while the normal force is 490 N and acts upwards.

Since the floor is horizontal, the vertical forces that act on the package are its weight and the normal force due to the ground.

The direction of the weight is downwards while the direction of the normal force is upwards.

Since the floor is horizontal and the package does not move in the vertical direction, the net vertical force is zero.

  • Let W = weight of package = mg where
  • m = mass of package = 50 kg and
  • g = acceleration due to gravity = 9.8 m/s² and
  • N = normal force

So, the net force N - W = 0

N = W

= mg

= 50 kg × 9.8 m/s²

= 490 N

So, the weight is 490 N and acts downwards while the normal force is 490 N and acts upwards.

<h3>c. Distance package moves</h3>

The package moves 1.8 m

Since distance, d = vt where

  • v = velocity = 0.6 m/s and
  • t = time = 3 s

So, d = vt

= 0.6 m/s × 3 s

= 1.8 m

So, the package moves 1.8 m

<h3>d. The package's acceleration</h3>

The package's acceleration is 0 m/s²

Since the net force on the package is zero, its acceleration is also zero. Since force, F = ma where

  • m = mass of package and
  • a = acceleration of package

Since F = 0,

ma = 0

a = 0

So, the package's acceleration is 0 m/s²

Learn more about force here:

brainly.com/question/25239010

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Answer:

The displacement of the ant, R = 29.15 cm

The angle of the resultant displacement with its original position is, θ = 30° 57'

The direction of the displacement is towards the northeast.  

Explanation:

Given data,

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The displacement towards  west, d₃ = 15 cm

The total displacement towards east,

                                       d₄ = d₁ - d₃

                                           = 30 - 15

                                           = 15 cm

The total displacement of ant is given by the resultant displacement,

                            R = √(d₂² + d₄² + 2· d₂ d₄ CosФ)

Where Ф is the angle between the vectors, d₂ & d₄

                                                Ф = 90°

Therefore,

                               R = √(d₂² + d₄²)

Substituting in the above equation,

                               R = √(25² + 15²)

                                  = 29.15 cm

Hence, the displacement of the ant, R = 29.15 cm

The angle of the resultant displacement with its original position is,

                                θ = tan⁻¹ (d₄ / d₂)

                                   = tan⁻¹ (15 / 25)

                                   = tan⁻¹ 0.6

                                   = 30° 57'

Hence, the angle of the resultant displacement the its original position is, θ = 30° 57'

The direction of the displacement is towards the northeast.                                            

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