What Causes Different Acceleration Calculations in Sled Dynamics?

[johndoe3344]

A boy drags his 60.0-N sled at constant speed up a 15 degree hill. He does so by pulling with a 25-N force on a rope attached to the sled. If the rope is inclined at 35 degrees to the horizontal,
(a) what is the coefficient of knetic friction between sled and snow?
(b) At the top of the hill, he jumps on the sled and slides down the hill. What is the magnitude of his acceleration down the slope?

I understand how to do part a. I get the correct answer: 0.161. However, I try to do part b, and I get a different answer than the back of the book. Here are my steps:

Fk = uk*n ; n = mg
Fk = uk*mg = ma
a = uk*g

I plug in the numbers and get 1.579 m/s/s, while the back of the book gets 1.01 m/s/s.

What am I doing wrong? Thanks.

 

[MalayInd]

Consider the boy and the sled as a system
There are three forces acting on the system,
their weight
force of friction
and, normal reaction.
Here n is not equal to mg but mgcos(15)
the net force along the incline is mgsin(15)-fk=mg(sin(15)-ukcos(15))
Hence the acceleration is"g(sin(15)-ukcos(15))"

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Malay