Free Body Diagrams & Net Forces practice

Choose the correct two answer choices of the objects forces acting on each other identified in the appropriate directions.

• The force of gravity of the table pulling down on the lamp

• The force of gravity of the lamp pulling down on the the table

• The normal force of the lamp pushing up on the desk

• the normal force of the desk pushing up on the lamp

Tags

NGSS.HS-PS2-1

Are the net forces of this object at equilibrium?

• yes, the Y component of the net forces are opposite in direction, so they cancel out. And the X component of the net forces are opposite in direction so they cancel out.

  $\Sigma F=0$

• No, the net forces add up to 14 N. $\Sigma F\ =\ 14N$

Tags

NGSS.HS-PS2-1

If this box's net forces are at equalibrium, is it possible that the box could still be in motion? 

• yes, if it was originally in motion when these forces were applied. Since the net force equals zero, there would be no change in motion. 

• no, a net force in equilibrium means 0, therefore, it is at rest. 

Tags

NGSS.HS-PS2-1

Assuming this object was originally at rest, will the net force cause an acceleration, (a change in its motion)? If so, in what direction will it accelerate?

• No, the net forces are balanced, therefore in equilibrium. The box remains at rest

• The forces are not balanced, so the Net force is not in equilibrium, so the box will accelerate to the left.

• The forces are not balanced, so the Net force is not in equilibrium, so the box will accelerate to the right.

• The forces are not balanced, so the Net force is not in equilibrium, so the box will accelerate up.

• The forces are not balanced, so the Net force is not in equilibrium, so the box will accelerate down.

Tags

NGSS.HS-PS2-1

What is the magnitude of the force causing the box to accelerate? What direction will it accelerate?

• $\Sigma F\ =\ 65N\ North$
  

• $\Sigma F\ =\ 15N\ North$

• $\Sigma F\ =\ 15N\ South$

• $\Sigma F\ =\ 0N\ \ @\ Equilibrium$

Tags

NGSS.HS-PS2-1

This box is originally traveling at a velocity of 25m/s out of the screen (z axis).

What is the magnitude of the net force acting on this box, and what direction will the net force cause the box to accelerate?

• $\Sigma F\ =\ 0N$ The box is at equilibrium. It will be at rest.

• $\Sigma F\ =\ 0N\$ It is at equilibrium. The box will remain traveling at 25 m/s in the Z axis direction out of the screen.

• $\Sigma Fy\ =\ 6N,\ \ \Sigma Fx=\ 10N$ The Net force is 16N.

Tags

NGSS.HS-PS2-1

Calculate the

 $\Sigma Fx$  

•  $\Sigma Fx\ =\ -8300\ N$  

•  $\Sigma Fx\ =\ 2300\ N$  

•  $\Sigma Fx\ =6000N$  

•  $\Sigma Fx\ =-6000N$  

•  $\Sigma Fx\ =\ 10,6000\ N$  

Tags

NGSS.HS-PS2-1

What is the magnitude of the normal force?

•  $Fnormal\ =\ -9500\ N$  

•  $Fnormal\ =\ 9500\ N$  

•  $Fnormal\ =\ 1000kg$  

Tags

NGSS.HS-PS2-1

What is the equation for the force of gravity?

• Fg=mg

• Fg=ma

Answer explanation

we need to be more specific about what type of acceleration we are using in the equation as there is the x component of acceleration, and there is the y component, which is the acceleration due to gravity. 

Tags

NGSS.HS-PS2-4

Calculate the force due to gravity vector component.

• -9800 N

• 9800 N

• 10500 N

• 8500 N

Answer explanation

Fg=mg,  Therefore..
Fg = (1000)(-9.8)
Fg= 9800 N

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-4