Help on Physics Q: M=400g, V0=80cm/s, F=.7N: A & B

  • Thread starter Thread starter PurposeDriven
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
The discussion revolves around solving a physics problem involving a 400g object sliding on a surface against a frictional force of 0.7N. For part A, the user initially calculated the distance the object slides before stopping but misused the average velocity and negative signs, leading to an incorrect distance of approximately 0.1829m. For part B, the calculation of the coefficient of kinetic friction was also flawed due to using an incorrect mass in the normal force calculation. Participants emphasized the importance of using the correct values and signs in equations to arrive at accurate results. Overall, the user is encouraged to double-check their calculations and apply the correct physics principles.
PurposeDriven
Messages
1
Reaction score
0
If anyone could help me on a physics question, I would sincerely appreciate it. I have already done it, but somehow I feel it's wrong . If someone could check it and tell me what I'm doing wrong if it is wrong, that would be great!

1) A 400g object with an initial speed of 80cm/s slides along a horizontal desktop against a frictional force of .7N.
A) How far will it slide before stopping?
B) What is the coefficient of friction between the object and the
table top?

So what I have so far is M=400g, Initial Velocity=80cm/s, Force of kinetic friction is .7N, Final Velocity is 0, and average velocity is 40cm/s. I know I first want to solve for d.

I used F=Ma ( .7N=400g(a) ) and found that acceleration is 1.75m/sec^2
I then used t=V(final)-V(initial)/a to solve for time. I converted my initial velocity into meters/sec. (-.8/1.75= -.4571428571) This makes sense that it's negative since it's deaccelerating.
Then I plugged that inThen I used V(av)=d/t to solve for d. (.4m/s*.4571428571=.1828571429) I dropped the negative b/c distance can't be negative. (Can I do that?) So my answer for A is .182857429m

Now for B I used the equation Mu sub k(coefficient of kinetic friction)=Force of Kinetic friction/Normal Force. So basically, Mu-sub-k=.7/(1.4*9.8) and ended up with .1785714286e-4.

Thank you so much!
 
Physics news on Phys.org
Originally posted by PurposeDriven
I used F=Ma ( .7N=400g(a) ) and found that acceleration is 1.75m/sec^2
I then used t=V(final)-V(initial)/a to solve for time. I converted my initial velocity into meters/sec. (-.8/1.75= -.4571428571) This makes sense that it's negative since it's deaccelerating.
Be more careful with your signs. The acceleration is negative, but the time better be positive!
Then I plugged that inThen I used V(av)=d/t to solve for d. (.4m/s*.4571428571=.1828571429) I dropped the negative b/c distance can't be negative. (Can I do that?) So my answer for A is .182857429m
Your general method (and your answer) is fine. If you are careful with signs, nothing will need to be "dropped". Also, don't use so many significant figures in your answer: three is plenty.

Of course, you did much uneeded work. If you are able to use energy methods, that's the way to go:
Workdone by friction = ΔKE
Now for B I used the equation Mu sub k(coefficient of kinetic friction)=Force of Kinetic friction/Normal Force. So basically, Mu-sub-k=.7/(1.4*9.8) and ended up with .1785714286e-4.
Your method is correct, but you used the wrong value for mass (0.4, not 1.4). Redo the calculation and you should get the right answer.
 



Hello, it looks like you are on the right track with your calculations. However, there are a few things to consider when solving this problem.

1) For part A, you correctly used the equation d=Vav*t to find the distance traveled. However, you used the average velocity as 40cm/s, which is incorrect. The average velocity is actually the initial velocity, since the object starts at 80cm/s and ends at 0cm/s. So the correct equation to use would be d=V0*t.

2) When you converted the initial velocity to meters per second, you used a negative sign. This is not necessary, as you are only calculating the magnitude of the velocity. So the correct value to use would be 0.8m/s, not -0.8m/s.

3) For part B, you correctly used the equation Mu-sub-k=Force of kinetic friction/Normal force. However, you used the incorrect value for the normal force. The normal force is the force exerted by the table on the object, which is equal to the weight of the object. So the correct equation to use would be Mu-sub-k=0.7/(0.4*9.8), which gives a value of 0.1785714286.

I hope this helps! Please double check your calculations and make sure to use the correct values in your equations. If you have any further questions, don't hesitate to ask. Good luck!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top