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 accelerates upward at a rate of 2.5m/s^2. what is the force exerted on the string?

Question

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 accelerates upward at a rate of 2.5m/s^2.  what is the force exerted on the string?

Fg=ma
=2*9.8
=19.6N

Fn=ma-Fg
=2*2.5-(-19.6)
=24.6N

b)she sets the box on the seat beside her. the bus acelerates from rest to 60 km/h in 4 s and the box begins to slide. What's the coefficient ofstatic friction b/w the box and te bus seat?

Ff=m*Fn
=2(24.6)
=49.2N

static coefficient=Ff/Fn
=49.2/24.6
=2

c)a taxi suddenly cuts infront of the bus, causing the bus driver to slam on the brakes. The bus driver reduces the speed from 60 km/h to 20 km/h in 1.5 s. Does Cathy's cake slide forward?
I have no idea how to even start with this one.

sarah · Answer

I am taking this course and am confused with this question... in the second part to(a),why does 19.6 become negative to get 24.6?
If it accelerating upward wouldn't 19.6 be positive making the answer
2(2.5)-19.6
=-14.6N
???