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

In summary, the cake box slides forward due to the coefficient of static friction between the box and the seat of the bus.
  • #1
ILC grade 11
5
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:
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  • #2
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.
 
  • #3
The accelaration of the bus is 4.2 m/s^2
 
  • #4
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?
 
  • #5
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
 
  • #6
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.
 

1. What is motion near earth's surface?

Motion near earth's surface refers to the movement of objects on or near the surface of the Earth. This includes both natural movements, such as the rotation of the Earth on its axis and the orbit of the Earth around the sun, as well as the movement of human-made objects, such as cars, planes, and satellites.

2. How is motion near earth's surface affected by gravity?

Gravity plays a major role in motion near earth's surface. It is the force that pulls objects towards the center of the Earth, causing them to fall towards the ground. This is why objects that are dropped from a height will accelerate towards the ground at a rate of 9.8 meters per second squared.

3. What factors affect the speed of motion near earth's surface?

The speed of motion near earth's surface is affected by several factors. These include the force of gravity, air resistance, and the mass and shape of the object. The presence of friction and inclines on the surface can also impact the speed of an object's motion.

4. How does air resistance affect motion near earth's surface?

Air resistance, also known as drag, is the force that opposes the motion of an object through the air. This force increases as the speed and surface area of the object increases. Therefore, air resistance can slow down the motion of objects near earth's surface, such as airplanes and cars, and can also affect the trajectory of falling objects.

5. How is motion near earth's surface measured?

Motion near earth's surface can be measured using various units of measurement. Common units include meters per second (m/s) for speed and acceleration, meters (m) for distance, and seconds (s) for time. Other units, such as miles per hour (mph) and feet (ft), may also be used depending on the context and location.

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