How Far Will a Puck Slide with Friction?

Click For Summary
A puck with an initial velocity of 7.5 m/s experiences a frictional force of 3.2 N while sliding on a horizontal surface. The calculations indicate that the puck will travel approximately 9.7 meters before coming to a stop. The initial attempt at solving the problem incorrectly applied constant velocity equations, leading to an overestimation of distance. The correct approach involves using the deceleration caused by friction to determine the stopping distance. Properly applying the equations of motion is essential for accurate results in physics problems.
Anuj Agarwal
Messages
2
Reaction score
0

Homework Statement


A force of friction of 3.2N acts on a 1.1Kg puck while it is sliding along a horizontal surface. If the initial velocity of the puck was 7.5 m/s, how far will the puck travel before coming to rest?

Homework Equations

The Attempt at a Solution


The answer is 9.7M. Tried multiple time with different formulas but unable to derive at solution.
 
Physics news on Phys.org
Hi @Anuj Agarwal, Welcome to Physics Forums.

You need to show us at least one of your attempts in detail or your thread will be deleted. Just saying you tried and failed does not give us any basis upon which to help you.
 
Sorry. Here is my try: F =mxa hence a = f/m=3.2/1.1=2.9
A=v/s, hence s=v/a =7.5/2.9=2.58
Distance = velocityxtime = 7.5x2.58=19.3 m
 
The last line is wrong. This would only be true if the velocity were constant. The fact that your answer is twice as big as it should be should give you a clue.
 
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
I treat this question as two cases of Doppler effect. (1) When the sound wave travels from bat to moth Speed of sound = 222 x 1.5 = 333 m/s Frequency received by moth: $$f_1=\frac{333+v}{333}\times 222$$ (2) When the sound wave is reflected from moth back to bat Frequency received by bat (moth as source and bat as observer): $$f_2=\frac{333}{333-v}\times f_1$$ $$230.3=\frac{333}{333-v}\times \frac{333+v}{333}\times 222$$ Solving this equation, I get ##v=6.1## m/s but the answer key is...