Help Week6 Begin Date: 7/7/2023 12:01:00 AM -- Due Date: 7/18/2023 11:59:00 PM End Date: 8/10/2023 11:59:00 PM constant value of 1550 N. (6%) Problem 4: Suppose a 1500-kg elevator car is lifted a distance 39 m by its cable at a constant speed, assuming the frictional force on the elevator has a 33% Part (a) Calculate the work done on the elevator by its cable, in joules. We = 633750 We = 6.338 x 105 Correct! 33% Part (b) Calculate the work done by the gravitational force on the elevator, in joules, during this process. Wg = - 573300 W. = -5.733 x 10" Correct! 33% Part (c) Calculate the total work done on the elevator, in joules. Wnet = 1 Grade Summary Deductions Potential 100 sin() cos() tan() 7 8 9 HOME Submissions cotan() asin() acos() E 4 5 6 Attempts remaining: 9 (0% per attempt) atan( acotan( sinh( 1 2 detailed view cosh() tanh() cotanh() + - 0 END O Degrees O Radians VO CLEAR Submit Hint I give up!(4%) Problem 7: The diagram shows a crate with mass m = 33.9 kg being pushed up an incline that makes an angle d = 15.7 with the horizontal. The pushing force is horizontal, with magnitude P, and the coefficient of kinetic friction between the crate and the incline is u = 0.37. Consider the work done on the crate as it moves a distance d = 5.73 m at constant speed. P Wp = 1331.62 25% Part (a) What is work, in joules, done by the pushing force, P? Wp = 1332 J V Correct! 25% Part (b) What is the work, in joules, done by friction? W/ = - 816.5 W/ = -816.5 J / Correct! 25% Part (c) What is the work, in joules, done by gravity? W. = - 515.12 W. = -515.1 J / Correct! $ 25% Part (d) What is the net work, in joules, done by the force P, friction and gravity? (Note that this is one approach to the problem. Either we consider the work done by gravity, or, more commonly, we will consider the change in gravitational potential energy.) Wnet = Grade Summary Deductions 0% Potential 100% sin() cos() tan( 7 8 9 Submissions cotan( 4 5 6 Attempts remaining: 994 asino acoso (0% per attempt atan() acotan() sinh( 1 2 3 detailed view 0% A cosho tanho cotanh() + 0 0% W I Degrees O Radians VO 0%WCCKo Digni Date. (6%) Problem 9: The force required to compress a non-standard spring as a function of displacement from equilibrium x is given by the equation F(x) = ax- - bx, where a = 65 N/m2, b = 18 N/m, and the positive x direction is in the compression direction of the spring. 4 33% Part (@)_Write a general equation in terms of the given variables for the work required to compress this spring from equilibrium to any point xp. W = Grade Summary Deductions 0% Potential 100% B a ( ) 7 8 9 HOME b d Submissions g 4 5 6 Attempts remaining: 999 h j K 1 2 3 (0% per attempt) detailed view m n P + - 0 Xp y VO CLEAR Submit Hint Feedback I give up ! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback. $3 33% Part (b) Calculate the work done, in joules, on the spring as it is compressed from .x = 0 to .x] = 42 cm. Wo--1 = 1.700 X Attempts Remain 33% Part (c) Calculate the work done, in joules, on the spring as it is compressed from .x] = 42 cm to x2 = 82 cm. W1-2 =5.873 - Correct!(8%) Problem 13: The person in the diagram pushes a lawn mower a distance d, applying a constant force of magnitude F that makes an angle 0 with the displacement, which is horizontal. A constant frictional force of and the normal force. magnitude facts opposite in direction to the displacement. The only other forces on the lawn mower are gravity d E 25% Part (a) Which of the forces do zero work on the lawn mower? Check all that apply. Gravity , The normal force Correct! 25% Part (b) Write an expression for the work WF done by the applied force in terms of F, d, and 0. WE = F d cos(0)