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3 years ago

A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates 1.5 m/s

Physics

2 answers:

Ronch [10]3 years ago

4 0

Answer:

Net force acting on the box is 75 N.

Explanation:

It is given that,

Mass of the box, m = 50 kg

Acceleration of the box, $a=1.5\ m/s^2$

Let F is the net force acting on the box. It can be calculated using the Newton's second law of motion. According to this law the force acting on an object is equal to the product of mass and acceleration with which it is moving. Mathematically, it is given by :

$F=m\times a$

$F=50\ kg\times 1.5\ m/s^2$

F = 75 N

So, the net force acting on the box is 75 N. Hence, this is the required solution.

Harrizon [31]3 years ago

3 0

We now that follow newton rules f=ma so net force equal to mass*acceleration=>f=50*1.5=75 N

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Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha

Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

$F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}$

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

$a_B = a_A-0.5 \frac{m}{s^2}$

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

$30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N$

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

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3 years ago

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A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w

Luden [163]

Here it is given that initial speed of the package will be same as speed of the helicopter

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displacement of the package as it is dropped on ground

$d = -105 m$

acceleration is due to gravity

$a = -9.8 m/s^2$

now by kinematics

$y = v* t + \frac{1}{2}at^2$

$-105 = 5.1 * t - \frac{1}{2}*9.8*t^2$

$4.9 t^2 - 5.1 t - 105 = 0$

by solving above equation we have

$t = 5.2 s$

so it will take 5.2 s to reach the ground

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Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

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final velocity of the sprinter, v = 27 km/h

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Convert the velocity of the sprinter to m/s;

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