Net Force and Friction Concepts

Net Force and Friction Concepts

Assessment

Interactive Video

Physics

9th - 10th Grade

Hard

Created by

Patricia Brown

FREE Resource

The video tutorial explains how to find the force of friction acting on an object using a force diagram and Newton's second law. It starts with a problem statement involving an applied force and acceleration, then proceeds to draw a force diagram. Known values such as mass and acceleration are identified, and the net force is calculated. Finally, the force of friction is determined by solving an equation involving the applied force and net force.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the applied force used to accelerate the object in the example?

130 Newtons

200 Newtons

170 Newtons

150 Newtons

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which force opposes the force of gravity in the force diagram?

Net force

Normal force

Applied force

Frictional force

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the mass of the object being accelerated?

10 kilograms

5 kilograms

8 kilograms

12 kilograms

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is the net force calculated in this example?

By subtracting mass from acceleration

By multiplying mass and acceleration

By dividing mass by acceleration

By adding mass and acceleration

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the net force acting on the object?

104 Newtons to the left

66 Newtons to the left

104 Newtons to the right

170 Newtons to the right

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What equation is used to find the force of friction?

Force of friction = Net force - Applied force

Force of friction = Applied force - Net force

Force of friction = Applied force / Net force

Force of friction = Applied force + Net force

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the calculated force of friction acting on the object?

104 Newtons to the left

66 Newtons to the right

66 Newtons to the left

170 Newtons to the right

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