# Motion near earth's surface

## 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.

## The Attempt at a Solution

I am done with A
but help with b and c will be really appreciated

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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

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.