If hand written please be neat! Thank you for your help! 0.20 and an (numbers 14
ID: 1789133 • Letter: I
Question
If hand written please be neat! Thank you for your help!
0.20 and an (numbers 14-18) A 5.0 kg block is on a horizontal surface with coefficient of static friction unknown coefficient of kinetic friction. A person pushes horizontally on the block. of magnitude 2.0 m/s. What 14. When the block is pushed with 8.0 N of force, it slides with a is the magnitude of the kinetic friction force? a) O N b) 5N c) 6N d) 8N e) 10 N Push 15. While still sliding, the pushing force is reduced such that the block now slows down at a constant rate of 1.0 m/s2. What is the magnitude of the applied pushing force? a) 1N b 2N c) 3 N d) 4N e) 5 N 16. The block is now allowed to come to rest. After coming to rest, a 12 N push is applied. The surface below applies a a) static, 10 b) static, 12 c) kinetic, 8 d) kinetic, 12 e) none of the above answers is correct friction force of magnitude Newtons. 17. An additional 5.0 kg block is now attached to the top of the original block and again, the composite block (10.0 kg total) begins at rest. An 18 N push is now applied. The surface below applies a force of magnitude a) static, 18 b) static, 20 c) kinetic, 16 d) kinetic, 18 e) none of the above answers is correct friction Newtons.Explanation / Answer
14) Since the block is moving with constant velocity,there is no acceleration and therefore the net force acting on block is zero.Friction force =Applied force=8N
15)The applied force is reduced such that block is decelerating at 1 m/s2
Let the applied force be F
F-8 = 5*(-1)=-5
F=8-5=3N
16)Since when applied force was 8N,the block was moving at constant velocity, the maximum kinetic friction force the ground can exert is 8N.Therefore, when applied force is 12N,the block will be sliding and the ground will exert kinetic friction of 8N on the block
17)Total mass of system=10 kg
Applied force=18 N
maximum static Friction force = 0.2*10*9.8=19.6N
since the applied force is less than maximum static friction force,the surface will apply a static friction force of 18N.(note: friction force cannot exceed applied force)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.