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Question

question1:
A 40.5 kg girl and a 58.6 kg boy are on the surface of a frozen lake, 13.0 m apart. Using a rope, the girl exerts a horizontal 4.30 N force on the boy, pulling him toward her. Calculate the magnitude of the girl's acceleration.

question 2 :
The coefficent of static friction between the floor of a truck and a box resting on it is 0.36. The truck is traveling at 78.6 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?

pls help meeeeeee

Answer 1

Answer 1:
There's a lot of extraneous information here. We don't need to know the boy's mass or the distance apart. All we need to know is the force exerted on the boy and the mass of the girl. Given that the girl exerts a 4.30N force on the boy, an equal and opposite force is exerted on the girl by the boy (Newton's third law) Thus the net force on the girl is 4.30N. Now use the equation F=ma )Newton's second Law) to solve for the acceleration.
4.30 N = (40.5kg)(a)
a = .106 m/s^2

Question 2: In this question, we will first find the maximum force of static friction on the box. Then we will find the maximum acceleration this friction can cause to stop the box. Then we will find how far the truck goes while (de)accelerating that much.

Ffriction = (FNormal)(coefficient of static friction)
Ffriction = (m)(9.8m/s^2)(0.36)
Ffriction = (3.53m/s^2)m

Now the acceleration, found by F = ma
(3.53 m/s^2)m = m(a)
a = 3.53m/s^2

Now, the initial velocity of the car is 78.6 km/hr = 21.8 m/s
Using the equation (final velocity)^2 = (initial velocity)^2 + 2(a)(d)
0 = (21.8 m/s)^2 + 2(-3.53 m/s^2)(d)
-475.24 = -7.06 d
d = 67.3 meters
The truck can stop in 67.3 meters without having the box slip.