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3 years ago

What is the Net Force acting on the box below?

Physics

1 answer:

Eddi Din [679]3 years ago

4 0

The Net Force would be 2 N to the left.

21 N is being used to push the box to the right and 23 N is used to push it left. There is a stronger force pushing the box towards the left. The different in the two numbers would give you the net force acting on the box and the direction of the arrow with the greatest force will tell you the direction.

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What is the magnitude and direction (right or left) of the

Sunny_sXe [5.5K]

Answer: 12 N to the right

Explanation:

If we calculate the net force acting on the box, we will have:

<u>In y-component:</u>

$Fy_{net}=F_{n}+F_{g}$ (1)

Where $F_{n}=12 N$ is the Normal force, directed upwards and $F_{g}=-12 N$ is the weight of the box (gravity force), directed downwards.

$Fy_{net}=12 N-12 N$ (2)

$Fy_{net}=0 N$ (3) Hence the net force in the vertical component is zero

<u>In x-component:</u>

$Fx_{net}=F_{left}+F_{right}$ (4)

Where $F_{left}=-3 N$ and $F_{right}= 15 N$

$Fx_{net}=-3 N + 15 N$ (5)

$Fx_{net}=12 N$ (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right

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2 years ago

What exercise became part of movement in the 1970

joja [24]

Answer:

Aerobic exercise became part of this movement in the 1970s.

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3 years ago

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.