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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 (6%) Problem 18: A
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 (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 point C is horizontal distance x from point B. The speed of the block at point B is VB- 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 =1 Grade Summary Deductions Potential B - Y () 7 8 9 HOME Submissions b d g 4 5 6 Attempts remaini 0% per attempt h K 1 2 3 detailed view m n P + - 1 0 S VB X VO Submit Hint I give up! Feedback: 0% deduction per feedback. Hints: 1 for a 0% deduction. Hints remaining: 0 According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. Three forces act on the block: The normal force, gravity, and kinetic friction. The work done by a constant force is given by W = FAr cost, where F is the magnitude of the force, Ar is the magnitude of the displacement, and 8 is the angle between the force and the displacement. The normal force does no work because it is perpendicular to velocity of the block at all times. Thus, the work done by gravity plus the work done by friction is equal to the change in the kinetic energy of the block.Class Management | Help theExpertTA.com 1 Student: telghazi2101@student.stcc.edu 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 21: Suppose a 775-kg car climbs a 2.050 slope at a constant 30.5 m/s. *50% Part (a) What is the power output of the car P, in terms of the variables given in the problem statement, if the friction and wind resistance encountered is P = (F+(mg )/( sin(0) )| Grade Summary Deductions 0% Potential 100% cos(a.) cos(Q) cos(0) ( ) 7 8 9 HOME Submissions sin(a sin(0) sin(0) 5 6 Attempts remaining: 987 (0% per attempt) B 3 detailed view d F g END m V VO BACKSPACE 10% DEL CLEAR 09% 0% Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 1 Feedback: 0% deduction per feedback. 50% Part (b) What is the power output of the car P, in watts, if F= 590 N? P = 26289.86 P = 2.629 x 104 Correct
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