Free-Body Diagrams, Friction, Action and Reaction

A box with a mass of 0.824 kg is being dragged across a rough horizontal surface with an 8.3 N force to the right. A friction force of 2.6 N to the left acts on the box. What is the net horizontal force on the box?

• 0 N

• 5.7 N left

• 10.9 N right

• 5.7 N right

Answer explanation

The net force is the combination of the forces.

F(net) = 8.3 N + -2.6 N

F(net) = 5.7 N

Positive means to the right

F(net) = 5.7 N right

Tags

NGSS.MS-PS2-2

The net force on the 1.15-kg box shown in the picture is 2.7 N to the right. The unknown friction force is about ---.

• 7.8 N right

• 5.1 N left

• 5.1 N right

• 11.2 N left

Answer explanation

The net force is the combination of the 7.8-N force to the right and the unknown force to the left.

F(net) = 7.8 N + F(f)

2.7 N = 7.8 N + F(f)

-5.1 N = F(f)

Tags

NGSS.MS-PS2-2

A stone is dropped, and as it falls through the air it encounters a drag force as shown in the free-body diagram. Calculate the net force on the stone.

• 800 N down

• 600 N down

• 400 N down

• 400 N up

Tags

NGSS.MS-PS2-2

Suppose the 5.8-kg crate shown in the picture is being pushed across a rough horizontal surface by an applied 34 N force to the right. As the box slides, a 25 N friction force to the left acts on the box. The box has an acceleration of about ---.

• 5.9 m/s²

• 1.6 m/s²

• 10.2 m/s²

• 0 m/s²

Answer explanation

The acceleration comes from Newton's Second Law of Motion:

F(net) = m·a

The net force is the combination of forces in the horizontal direction

F(net) = 34 N + -25 N

F(net) = 9 N

Positive means to the right

9 N = 5.8 kg·a

1.6 m/s² = a

Tags

NGSS.MS-PS2-2

If this plate of fries weighs 1.2 N, what mass of fries is being served? (Use g = 10 m/s² down)

• 120 kg

• 0.120 kg

• 8.8 kg

A 3.7-kg box is being pulled across a horizontal surface, as shown. The box accelerates at 1.1 m/s² to the right. The friction force on the box is about ---.

• 1.1 N left

• 4 N left

• 11 N left

• 36 N left

Answer explanation

The 15 N force and the friction force must combine to be the net force to the right.

The net force can be found from F(net) = m·a

F(net) = 3.7·1.1

F(net) = 4.1 N

So the two forces combining must look like

F(net) = 15 N + F(f)

4.1 N = 15 N + F(f)

-11 N = F(f)

Tags

NGSS.MS-PS2-2

What is happening to the rocket in the picture?

• It is being accelerated down

• It is moving down

• It is moving up

• It is being accelerated up

Tags

NGSS.MS-PS2-2

Rougher surfaces have ...

• greater friction
• less friction
• the same level of friction

When is having less friction useful?

• walking

• driving

• ice skating

• climbing

A book rests on a table.  The force of gravity pulls down on the book with a force of 20 newtons. What prevents the books from accelerating downward ?   

• The table presses back up on the book with an equal and opposite force of 20 newtons.

• The table presses back up with a force greater than 20 Newtons

• The tale and the book together are accelerating downwards at 9.8 m/sec2

• The inertia of the book holds it up

Tags

NGSS.MS-PS2-2