Class Management | Help the ExpertTA.com Student: telghazi2101@student.stcc 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. Wc = 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 WE = -5.733 x 105 Correct! 33% Part (c) Calculate the total work done on the elevator, in joules. W Wnet=1 Grade Summary Deductions 0% Potential 100% sin() cos() tan() TE ( 7 8 9 HOME Submissions 4 5 6 Attempts remaining: 995 cotan() asin() acos() (0% per attempt) acotan() sinh() detailed view atan() 1 2 3 0% cosh() tanh() cotanh() + 1 0 END 0% O Degrees O Radians VO CLEAR 10% 0% Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedback.Class Management | 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 (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 us ted ted ted Wp = 1331.62 25% Part (a) What is work, in joules, done by the pushing force, P? Wp = 1332 J Correct! ted ted 25% Part (b) What is the work, in joules, done by friction? W/=- 816.5 W/ = -816.5 J - Correct! ted ed 25% Part (c) What is the work, in joules, done by gravity? ed Wg = - 515.12 ed = -515.1 J V Correct! ed * 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.) ed ed Wnet Grade Summary Deductions 0% Potential 100% ed sin() cos() tan() TE ( 1 7 8 9 Submissions cotan() asin() acos() E 4 5 6 Attempts remaining: 994 (0% per attempt) atan() acotan() sinh( 1 2 3 detailed view ed cosh( tanh() 0 1 cotanh() 0% 4 0% Degrees O Radians VO A WN 0% 10% Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedbackWeeko 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 (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 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 Q. WE =F d cos(0) 25% Part (c) Write an expression for the work W- done by the frictional force in terms of fand d. Wf = Grade Sum Deductions Potential cos(0) sin(0) B () 7 8 9 HOME Submissions d 4 5 6 Attempts ren f F g 1 3 0% per atter detailed view h k 0 m P VO Submit Hint Feedback I give up! Hints: 1 for a 0% deduction. Hints remaining: 0 Feedback: 0% deduction per feedback -The work done by a constant force is given by W= Fdcoso, where F is the magnitude of the force, d is the magnitude of the displacement, and @ is the angle between the force and the displacement. 25% Part (d) Under what condition will the kinetic energy of the lawn mower increase? WF + Wf > 0 Correct! All content @ 2023 Expert TA, LLC #10Weeko 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 (4%) Problem 16: A 65-kg skydiver jumps out of an airplane and falls 490 m, reaching a maximum speed of 56 m/s before opening her parachute. Randomized Variables m = 65 kg h = 490 m v = 56 m/s 50% Part (a) How much work, in joules, did air resistance do on the skydiver before she opened her parachute? W = Grade Summary Deductions Potential sin() cos() tan( ) TE ( ) 7 8 9 Submissions cotan() asin( acos( ) E 4 5 6 Attempts remain (0% per attempt atan() acotan() sinh() 1 2 3 detailed view cosh( tanh() cotanh() 0 END N O Degrees O Radians VOl Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback. 50% Part (b) Assuming the force of air resistance is constant, what is the magnitude of the force of air resistance, in newtons, on her? Fave = 429.65 Fave = 429.7 Correct! All content @ 2023 Expert TA, LLCWeeko 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 (6%) Problem 18: A block with mass m is released from rest at point A and slides down a curved slope to point B, then continues until it comes to rest at point C. The vertical height of A is h above point B, and A point C is horizontal distance x from point B. The speed of the block at point B is vg. - 5 B C x * 50% Part (a) Write an expression for the work done by friction as the block slides from A to B. Your answer should be in terms of m, h, VB, and g (the acceleration due to gravity). W = Grade Summary Deductions 0% Potential 100% B Y ( 7 8 9 HOME Submissions b d g 4 5 6 Attempts remaining: 991 (0% per attempt) h k 1 2 3 detailed view m n P 0 1 0% 0% S VB X Vol CLEAR 0% 0% Submit Hint Feedback I give up ! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 0 % deduction per feedback. 50% Part (b) Write an expression for the coefficient of kinetic friction between the block and the horizontal surface. Your answer should be in terms of x, VB and g (the acceleration due to gravity). HK = VB (2gx)