A 6 kg box is sitting at rest. It has a coefficient of static friction of 0.5 and of kinetic friction of 0.2
I push on it with 15 N. What is the acceleration of the box, m/s2?
Friction and Ramps Quizizz
Quiz
•
Physics
•
12th Grade
•
Medium
KELLY DOYLE
Used 2+ times
FREE Resource
27 questions
Show all answers
1.
MATH RESPONSE QUESTION
2 mins • 1 pt
A 6 kg box is sitting at rest. It has a coefficient of static friction of 0.5 and of kinetic friction of 0.2
I push on it with 15 N. What is the acceleration of the box, m/s2?
Mathematical Equivalence
ON
Answer explanation
The maximum static friction is 30 N (normal force of 60 N multiplied by 0.5).
We did not push hard enough to break static friction, so the acceleration and the velocity are zero.
2.
MATH RESPONSE QUESTION
2 mins • 1 pt
A 8 kg box is sitting at rest. It has a coefficient of static friction of 0.4 and of kinetic friction of 0.2
I push on it with 40 N. What is the acceleration of the box, in m/s2?
Mathematical Equivalence
ON
Answer explanation
The maximum static friction is 32 N (normal force of 80 N multiplied by 0.4).
We pushed hard enough to break static friction. SO the box will experience kinetic friction instead. Kinetic friction is 16 N (normal of 80 multiplied by 0.2).
So, if we have a push of 40 N and an opposing kinetic friction of 16 N, our next force is 24 N. Divide that by mass (8 kg) to get 3 m/s2.
3.
MATH RESPONSE QUESTION
2 mins • 1 pt
A 12 kg box is sitting at rest. It has a coefficient of static friction of 0.3 and of kinetic friction of 0.1
I push on it with 20 N. What is the acceleration of the box, in m/s2?
Mathematical Equivalence
ON
Answer explanation
The maximum static friction is 36 N (normal force of 80 N multiplied by 0.4).
We did not push hard enough to make the box move (acceleration and velocity remain zero).
4.
MATH RESPONSE QUESTION
2 mins • 1 pt
A 12 kg box is sitting at rest. It has a coefficient of static friction of 0.3 and of kinetic friction of 0.1
I push on it with 60 N. What is the acceleration of the box, in m/s2?
Mathematical Equivalence
ON
Answer explanation
The maximum static friction is 36 N (normal force of 80 N multiplied by 0.4).
We broke static friction. Recalculate with kinetic friction!
Kinetic friction is 120 N (normal) multiplied by 0.1 (coefficient of kinetic friction) = 12 N
A push of 60 N and opposing friction 12 N results in 48 N as a net force. Divide that by mass for an acceleration of 4 m/s2.
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
A box is being pushed across a rough surface and is traveling at a constant speed. What must be true?
It is incurring static friction
Net force is zero
Kinetic friction is greater than applied force
Applied force is greater than kinetic friction
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
If the net force is zero, then the object's velocity is...
must be zero
decreasing
increasing
not changing
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
A box is sliding up a rough ramp. What direction does friction point?
Down the ramp
Up the ramp
Perpendicular to the ramp
Both up and down the ramp at the same time
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