You and your brother argue often about how to safely secure a toddler in a moving car. You insist that special toddler seats are critical in improving the chances of a toddler surviving a crash. Your brother claims that, as long as his wife is buckled in next to him with a seat belt while he drives, she can hold onto their toddler on her lap in a crash. You decide to perform a calculation to try to convince your brother. Consider a hypothetical collision in which the 14 kg toddler and his parents are riding in a car traveling at 47 mi/h relative to the ground. The car strikes a wall, tree, or another car, and is brought to rest in 0.13 s. You wish to demonstrate to your brother the magnitude of the force necessary for his wife to hold onto their child during the collision. What is the magnitude of this force (in N)?   ______________N

Mariana and Jun are both pushing on a box. Mariana pushes the box to the right and Jun pushes to the left. The box is on rollers and so is essentially frictionless with the floor, but Mariana and Jun do have friction with the floor. Jun is "winning" the pushing match and all three objects, Jun, the box, and Mariana are accelerating to the left. Consider the following forces: fm: the force of friction on Mariana f;: the force of friction on Jun FMB: the force of Mariana on the box FJB: the force of Jun on the box Which of the listed forces must be equal in magnitude and opposite in direction? Of; and FMB Mariana box Jun fm and FMB fm and fj no two of the four listed forces must be equal and opposite. FMB and FJB

It may appear at first glance that Newton's 2nd and 3rd laws are independent claims. This, however, is not true - Newton's 3rd law is simply a consequence of Newton's 2nd law. In this problem, you will prove Newton's 3rd law for the normal contact force between two objects, using only Newton's 2nd law. Consider two blocks with masses m₁ and m₂ in contact, with external forces F₁ and F2 applied, as shown in the figure. The free body diagram for each block is also given, with internal forces fa and f, as shown. F₁ F₁ fa fb ΣF = ΣF = m₁ m2 Submit F₂ Complete the equations of motion for the entire system, as well as blocks 1 and 2 individually. Note, the forces have a signed magnitude, where positive values correspond to the positive x-direction ΣF= = (m₁ + m₂) a = m₁ a = m₂ a Use the above equations to write fa in terms of only fr: fa = m₁ You have used 0 of 4 attempts m₂ F₂ Save Calculator

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