Physics 1 Problem Set 5 (2)

pdf

School

SUNY Empire State College *

*We aren’t endorsed by this school

Course

Subject

Aerospace Engineering

Date

Oct 30, 2023

Type

pdf

Pages

Uploaded by BailiffPencilAardvark36

Problem Set #5 – Problems 12, 20, 32, 42, 50 12. Driving home from school one day, you spot a ball rolling out into the street. You brake for 1.20 s, slowing your 950-kg car from 16.0 m/s to 9.50 m/s. (a) What was the average force exerted on your car during braking? (b) How far did you travel while braking? a. F = ma = m( ∆v/∆ t) = 950kg ( (9.50-16.0) / (1.20) ) = (950) (-5.42) = -5146 N = -5.1kN b. ∆x= 16.0 m/s (1.20 s) + ½ (-5.42 m/s^2) (1.20 s)^2 19.2+ -3.9024 = 15.297m = 15.3m 20. On vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85 m/s 2 . (a) What is the net force exerted on the trailer? (b) What force does the trailer exert on the car? (c) What is the net force acting on the car? a. F= ma =(560) (1.85) = 1036 N b. This force the trailer exerts on the car is equal to the force exerted on the trailer from part a: = 1036N c. F= ma = (1400) (1.85) = 2590N 32. A 65-kg skier speeds down a trail. The surface is smooth and inclined at an angle of 22° with the horizontal. (a) Find the direction and magnitude of the net force acting on the skier. (b)

Does the net force exerted on the skier increase, decrease, or stay the same as the slope becomes steeper? Explain. a. F= mg sin θ = 65 (9.81) (sin 22) = 238.87 N = .24kN b. The net force that was exerted on the skier would increase as the slope became more steep. This happens because as the angle increases, the slope also becomes more steep. 42. When you weigh yourself on good old terra firma (solid ground), your weight is 142 lb. In an elevator your apparent weight is 121 lb. What are the direction and magnitude of the elevator’s acceleration? Direction of elevators acceleration: downwards Magnitude: ΣF= ma ma= 121-142 = -21 lbs m= 142/g m= 142/ 9.81 = |a|= = | (-21 / 142) (9.81) = |1.45m/s^2| = 1.5m/s^2 50. A 5.0-kg bag of potatoes sits on the bottom of a stationary shopping cart. (a) Sketch a free-body diagram for the bag of potatoes. (b) Now suppose the cart moves with a constant velocity. How does this affect your free-body diagram? Explain.

a. b. If the cart was to move with constant velocity, the free body diagram would not change because constant velocity = zero acceleration of the cart. This would have no affect on the potatoes, making the diagram not change.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version