The feet of a standing person of mass m exert a force equal to mg on the floor, and the floor exerts an equal and opposite force upwards on the feet, which we call the normal force. During the extension phase of a vertical jump (see page 145), the feet exert a force on the floor that is greater than mg, so the normal force is greater than mg. As you learned in Chapter 4, we can use this result and Newton’s second law to calculate the acceleration of the jumper: a = Fnet/m = (n - mg)/m.Using energy ideas, we know that work is performed on the jumper to give him or her kinetic energy. But the normal force can’t perform any work here, because the feet don’t undergo any displacement. How is energy transferred to the jumper?

Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here F, = 437 N and F, = 359 N and friction is negligible. In the diagram below, the mass of the car = 3500 kg, e, = -25° and e, = 12°. (Assume the car faces the positive x-axis before the forces are applied.) F, (a) Find the resultant force (in N) exerted on the car. magnitude N direction (counterclockwise from the +x-axis) (b) What is the acceleration (in m/s?) of the car? magnitude m/s² direction (counterclockwise from the +x-axis)

A 1,500-N crate is being pushed across a level floor at a constant speed by a force F of 370 N at an angle of 20.0° below the horizontal, as shown in the figure a below. F 20.0° / b 20.0° (a) What is the coefficient of kinetic friction between the crate and the floor? (Enter your answer to at least three decimal places.) 0.220 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.

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