College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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A 9.3-kg box slides across the floor of an elevator with a coefficient of kinetic friction equal to 0.38. (a) What is
the
rest? (b) If the elevator moves upward with constant acceleration, is
the force of kinetic friction acting on the box greater than, less than,
or equal to the value found in part (a)? Explain. (c) Determine the
force of kinetic friction acting on the sliding box when the elevator
moves upward with a constant acceleration of 1.4 m>s
2
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- A 6.7 kg mass is released from rets while on an incline of 53.96 degrees. If the coefficient of kinetic friction is 0.478 what is the magnitude of the acceleration of the mass?arrow_forwardA 9.08 kg mass is released from rest while on and incline of 58.65 degrees. If the coefficient of kinetic friction is 0.301, what is the magnitude of the acceleration of the mass?arrow_forwardA daredevil bicyclıst moves at 7.0m/s in a horizontal cirde in a cylindrical "weli of dea th" of radius 4.0 m. (A.) Draw the free - bady diagram (B) What is the minimum coefficient of friction required ? man riding a bikearrow_forward
- Write Newton's laws for a static system (1) EF, = mg sin 0 – µ̟n = 0 in component form. The gravity force has two components. (2) EF, = n - mg cos 0 = 0 Rearrange Equation (2) to get an n = mg cos e expression for the normal force n. Substitute the expression for n into EF, = mg sin 0 – µ̟mgcos 0 = 0 → tan 0 =µ, Equation (1) and solve for tan 0. Apply the inverse tangent function to tan 0 = 0.350 → 0 = tan 1 (0.350) = 19.3° get the answer. LEARN MORE REMARKS It's interesting that the final result depends only on the coefficient of static friction. Notice also how similar Equations (1) and (2) are to the equations developed in previous problems. Recognizing such patterns is key to solving problems successfully. QUESTION A larger static friction constant would result in a: (Select all that apply.) O larger component of normal force at the maximum angle. O larger component of gravitational force along the ramp at the maximum angle. smaller component of gravitational force along the ramp…arrow_forwardThe glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to vertical walls and even to ceilings.arrow_forwardA crate of mass 17.0 kg rests on a level surface. If the coefficient of kinetic friction between the crate and surface is 0.440, calculate the following. edi ume (a) Calculate the normal force. (Give the magnitude.) (b) Calculate the magnitude of the kinetic friction force when a horizontal applied force of 91.0 N moves the crate. (c) Calculate the normal force when the 91.0-N applied force is exerted at an angle of 32.5 above the horizontal. (Give the magnitude.) (d) Calculate the magnitude of the kinetic friction force when the 91.0-N applied force is exerted at an angle of 32.5 above the horizontal.arrow_forward
- During strenuous exercise, it is possible to exert forces to the joints that are easily ten times greater than the weight being supported. What is the maximum force of friction (in N) under such conditions? (The frictional forces in joints are relatively small in all circumstances except when the joints deteriorate, such as from injury or arthritis. Increased frictional forces can cause further damage and pain.) she exert 67.0 kg of her mass on that knee?arrow_forwardTwo boxes (Box A = 9.7 kg and Box B = 5.4 kg) are connected by acord running over a pulley. The coefficient of kinetic friction betweenbox A and the table is 0.18. We ignore the mass of the cord and pulleyand any friction in the pulley, which means we can assume that a forceapplied to one end of the cord will have the same magnitude at theother end. We wish to find the tension of the cord while accelerating (inN), assuming the cord doesn’t stretch. As box B moves down, box Amoves to the right. Hint: Solve for the acceleration of the system first.arrow_forward
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