physics question on force and acceleration... need help!!?

Question

1. Suppose a classmate of yours has missed lecture for the last week. Write a full solution of the following problem so that your classmate can use it to help catch up with the material. you need to explain in detail each step of the precess you use to get to an answer.

a sled weighing 100.0N is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is 0.2. A penquin weighing 70.0N rides on the sled, as in the figure.
(a) to start with, the penquin digs in his claws so he is firmly attached to the sled. What value of F do you need for the sled and penguin to move at constant speed?
(b) after a while, our penguin gets tired of holding on with his claws. Now the coefficient of static friction between penguin and sled is 0.7, find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.

Answer 1

a) Weight of sled + penguin = 100 + 70 = 170 N
Since the coefficient of friction = 0.2, the force of sliding friction is this coefficient * the normal force on the ice
= 0.2 * 170 = 34 N

b) The normal force of the penguin on the sled is 70 N, so the maximum force that can be applied to the penguin (due to acceleration) without causing penguin/sled sliding is
= 0.7 * 70 = 49 N.

Note that this value exceeds the force required to keep the sled moving. In fact, when the sled is just sliding along, the net force is 0. So, to get a net of 49 N on the penguin, note that this implies an acceleration of
a = 49/(70/g) = 7g/10
which means that the net force on the sled + penguin must be
F = 170(7g/10) = 119g = 119*9.8 ) = 1166.2 N

But this has to be the total, after overcoming the previously calculated force required to just keep moving: 34 N

Therefore, the total force exerted on the sled must be:
1166.2 + 34 = 1200.2 N.

Answer 2

Frictional force problems have a common element. The frictional force is equal to the normal force times the coefficient of friction.

For A, the sled and penguin move as a single mass, and the sled is moving at a constant speed so a=0. Since F=m*a, then F=0

Sum the forces
The tension on the cord in the horizontal direction is
equal to the frictional force:
(100+70)*.2=tension

b) This type of question considers the force at the verge of slipping. The normal force of the penguin is 70N, so the maximum force the penguin can have exerted is 70*.7

Now, since there is acceleration of the system
70*.7=a*70/g
or a=.7*g
To get the tension of the cord in the horizontal
.7*g*(170)/g=tension-170*.2
or
tension=170*(.2+.7)
=170*.9

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