The possible magnitude for the force of static friction on the stationary cart is 72.1 N.
The given parameters:
- <em>Applied force on the cart, F = 72.1 N</em>
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Based on Newton's second law of motion, the force applied to object is directly proportional to the product of mass and acceleration of the object.
F = ma
Static frictional force is the force resisting the motion of an object at rest.

where;
is the frictional force

Thus, the possible magnitude for the force of static friction on the stationary cart is 72.1 N.
Learn more about Newton's second law of motion: brainly.com/question/25307325
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Answer:
v_f = 0.87 m/s
Explanation:
We are given;
F_avg = -17700 N (negative because it's backward)
m = 117 kg
Δt = 5.50 × 10^(−2) s
v_i = 7.45 m/s
Now, formula for impulse is given by;
I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s
From impulse momentum theory, we know that;
Change in momentum of particle is equal to impulse.
Thus,
Δp = I = m•v_f - m•v_i
Thus,
-973.5= 117(v_f - 7.45)
Thus,
-973.5/117 = (v_f - 7.45)
-8.3205 + 7.45 = v_f
v_f = - 0.87 m/s
We'll take absolute value as;
v_f = 0.87 m/s