Calculating Forces and Friction in Motion: A Cake on a Bus

  • Thread starter Thread starter ILC grade 11
  • Start date Start date
  • Tags Tags
    Motion Surface
Click For Summary
SUMMARY

The discussion focuses on calculating forces and friction related to a 2 kg cake box on a bus. For part (a), the force exerted on the string is determined using Newton's second law, resulting in a force of 5 N. In part (b), the bus accelerates from rest to 60 km/h in 4.0 seconds, yielding an acceleration of 4.2 m/s², which is crucial for calculating the coefficient of static friction. Part (c) examines whether the cake slides forward when the bus decelerates from 60 km/h to 20 km/h in 1.5 seconds, requiring an understanding of relative acceleration and frictional forces.

PREREQUISITES
  • Newton's Second Law of Motion
  • Conversion of units from km/h to m/s
  • Understanding of static friction and its coefficient
  • Free body diagrams for visualizing forces
NEXT STEPS
  • Calculate the coefficient of static friction using the formula μ = f/N
  • Learn about free body diagrams in physics to analyze forces
  • Study the effects of deceleration on objects in motion
  • Explore the relationship between acceleration and force in different contexts
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and forces, as well as educators looking for practical examples of motion and friction in real-life scenarios.

ILC grade 11
Messages
5
Reaction score
0

Homework Statement


25a) Cathy takes the bus home from work. In her hand she holds a 2 kg cake box tied together with a string. As she ascends the steps into the bus, the box accelarates at a rate of 2.5 m/s^2. What is the force exerted on the sting?
b) Cathy sets the box on the seat beside her. The bus accelarates from rest to 60 km/h in 4.0 s, and then the box begins to slide. What is the coefficient of static friction between the box and the seat of the bus.
c) A taxi suddenly cuts in front of the bus causing the bus driver to slam the brakes. The bus driver reduces speed from 60.0 km/h to 20 km/h in 1.5s. Does cathy's cake slide forward.


Homework Equations





The Attempt at a Solution


I am done with A
but help with b and c will be really appreciated
 
Last edited:
Physics news on Phys.org
b)find acceleration
after you found it,place a free body diagram on the box.
You can answer these in two ways, did you learn D'Amblert principle?
if Not:
What is the acceleration relative to the ground of the box before it slides? what force makes it accelerate relative to the ground?
Good luck.
 
The accelaration of the bus is 4.2 m/s^2
 
Show me how You've calculated it please.
remember going from kilometers/h to meters/h you should divide by 3.6.
then a= (vf-vi)/(t1-t0)
and you got 4.2??
Anyway, how can You relate the acceleration relative to the Earth with the force acting on it relative to the earth,using which law?
 
a = v2-v1/ t
a= 16.7 m/s - o.o m/s / 4.0s
a= 4.2 m/s^2
um i converted the km into m/s which were 60 km and 0 km
 
ILC grade 11 said:
a = v2-v1/ t
a= 16.7 m/s - o.o m/s / 4.0s
a= 4.2 m/s^2
um i converted the km into m/s which were 60 km and 0 km

Didn't get where the 16.7 came from, now i understand
in b the bus accelerates from 0 to 20 km\h not to 60 km\h.
so You divided 60 by 3.6 and not 20 as in b.
anyway continue, relate the acceleration with the cake and the force that acts on it, tell me if You're having trouble.
 

Similar threads

Replies
9
Views
5K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
6
Views
14K
  • · Replies 2 ·
Replies
2
Views
7K