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# a 6.00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface as shown in the figure. the 6.00-kg block is being pushed by a horizontal 20.0-n force as shown. what is the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block?

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get a 6.00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface as shown in the figure. the 6.00-kg block is being pushed by a horizontal 20.0-n force as shown. what is the magnitude of the force that the 6.00-kg block exerts on the 4.00-kg block? from EN Bilgi.

## Solved A 6.00

Answer to Solved A 6.00-kg block is in contact with a 4.00-kg block on Source : www.chegg.com

## A 6.00

Answer to: A 6.00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface. The 6.00-kg block is being pushed by a...

Newton's laws of motion

## A 6.00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface. The...

A 6.00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface. The... Question:

A 6.00-kg block is in contact with a 4.00-kg block on a horizontal frictionless surface. The 6.00-kg block is being pushed by a horizontal 20.0-N force.

A) Draw the free-body diagram and write the equation of motion for the 6.0-kg block.

B) Draw the free-body diagram and write the equation of motion for the 4.0-kg block.

C) What is the common acceleration of the blocks?

D) What is the magnitude of the force that the 6.00 kg block exerts on the 4.00-kg block?

## Newton's Second Law of Motion:

Suppose two toy cars are connected to each other by using inelastic wire, and if we pull a toy car, of course, both toy cars start accelerating at the same rate. This acceleration is known as common acceleration, and to provide the same to each toy car, the net force acting on each toy car is directly proportional to the mass of that toy car. Mathematically, we can write:

∑ → F = m → a ∑F→=ma→ where: → F F→ is the force → a a→ is the acceleration m m is the mass

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Given data:

m 1 = 6.00 k g m1=6.00 kg

is the mass of the block

1 1 . m 2 = 4.00 k g m2=4.00 kg

is the mass of the block...

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## A 6.00 kg block is in contact with a 4.00 kg block on a horizontal frictionless surface. The 6.00 kg block is being pushed by a horizontal 20.0 N force. What is the magnitude of the force that the 6.00 kg block exerts on the 4.00 kg block?

Answer (1 of 4): Since the 6 kg block continuously pushes the 4 kg block, both accelerate at the same rate. The acceleration a is: a = \dfrac{F}{m} = \dfrac{20}{(6 + 4)} = 2 m/s^2 Therefore the force F exerted by the 6 kg block on the 4 kg block is: F = ma = (4 kg)(2 m/s^2) = 8 N A 6.00 kg block is in contact with a 4.00 kg block on a horizontal frictionless surface. The 6.00 kg block is being pushed by a horizontal 20.0 N force. What is the magnitude of the force that the 6.00 kg block exerts on the 4.00 kg block?

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, Ph.D. Physics, University of Alberta (1973)

Since the 6 kg block continuously pushes the 4 kg block, both accelerate at the same rate. The acceleration

a a is: a= F m = 20 (6+4) a=Fm=20(6+4) =2 =2 m/s 2 2 Therefore the force F F

exerted by the 6 kg block on the 4 kg block is:

F=ma=(4 F=ma=(4 kg )(2 )(2 m/s 2 )=8 2)=8 N

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How much force does the 4 kg block exert on the 6 kg block?

Two blocks (a and b) are in contact on a horizontal frictionless surface. A 60 N constant horizontal force is applied to A. The mass of a is 3.0 kg and the mass of b is 15 kg. what is the magnitude of the force of a on b?

Blocks with masses of 5.0 kg, 8.0 kg, and 9.0 kg are lined up in a row on a frictionless table. All three are pushed forward by a 20 N force applied to the 5.0 kg block. How much force does the 8.0 kg block exert on the 5.0 kg block?

A 70 kg man pushes a 50 kg man with a force of 50 N. With what force is the person pushed?

A 10 kg block lies on a smooth ramp that is inclined at 30°. What force, parallel to the ramp, would prevent the block from moving (assume that 1 kg exerts a force of 9.8 N)?

Scott Korba

, MBA, Academic Fellow, from University of North Carolina at Chapel Hill (1996)

Assuming the force is applied along the perpendicular tangent to the abutting block faces (so the force is acting against a 10KG ‘block'), then the force being applied, by definition, is divisible by the relative masses. 60% of the force is 'consumed' by the 6kg block and 40% by the 4 kg block. They will accelerate together at the same rate without any gap between them.

12N moves the 6Kg block and 8N moves the 4Kg block. They'll accelerate at 2m/sec for the duration the force is applied. (20N/10Kg)

Semantic argument about one block exerting a force against another since they are touching and ass

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Oscar Villalobos

, studied at San Francisco State University

Every action has an opposite and equal reaction applies to the point of contact in between blocks.

Friction isn't applicable because the point of contact between blocks is only taken into account.

This means horizontal force F1=F2= 20 N.

1.7K views Xuan Bach Lai Answered 1 year ago

Weight doesn't matter in an ideal frictional surface, and the question mentioning “frictionless” seems like an ideal case to me.

Since you didn't specify, there are 2 sets of friction coefficient, block vs block (coefficient BB) and block vs resting surface (coefficient BS), which one is frictionless? Both? Anyways, the resulting force of block-6kg on block-4kg would be 20 N * BB in the same direction as the applied 20 N. It's the reaction force of the same magnitude (Newton's law) against the friction force that block-4kg exerts on block-6kg. If BB is zero (no friction) the result would be zer

A car weighs 15300 N. What is its mass?

This seems like a simple problem, but it lacks additional, crucial data to solve it. Is may seem funny but it is important to know: Is the car on the surface of the Earth? Or is it on the surface of the Moon? Because using Newton’s Universal Gravitation Law, since Earth is bigger than the Moon it will generate a stronger Gravitational field, hence a higher value of gravtiational acceleration.

If we take:

1)The car to be on the surface of the Earth

then the value of g, would be approximately 9.81 m/s^2

Using the formula G=mg

We can arithmetically find that the mass of the car is nearly 1560 kilogram

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