Solving Physics Questions: Acceleration & Friction Coefficient

  • Thread starter Thread starter staz13
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
To determine the acceleration of a cart rolling down a 27-degree ramp with frictionless wheels, the formula F=ma is applied, leading to an acceleration of approximately 4.45 m/s² using the equation a = g sin(27). For the second question regarding the coefficient of friction between a coin and a book, the force of friction is expressed as f = uN, where N is the normal force, calculated as mg cos(25). The relationship between the forces is established through f = mg sin(25), allowing for the calculation of the coefficient of friction (u) by rearranging the equations. The discussions clarify the application of gravitational forces in both scenarios, emphasizing the importance of understanding normal and frictional forces in physics problems.
staz13
Messages
2
Reaction score
0
Hey guys! I REALLY need some help with my physics. Here is one of my two questions:

1) With what rate of acceleration will a cart roll down a ramp inclined at an angle of 27 degrees? Assume the wheels of the cart are frictionless

I have NO idea how to even get started.

Here is the second one, 2) A coin is placed on a book that is slowly opened. It is noted that the coin does not start to move until the opening angle is 25 degrees. What is the coeficient of friction between the book and the coin?

THANKYOU in advance if you can help at all!
 
Physics news on Phys.org
1.use F=ma for the first one. So mgxsinx27=ma, then gxsin27=a

2. Use f=un, so mgcos25=Normal.

Maybe use F-f=ma to find f then use f=un to find coefficient of friction
 
For question 1 I got 4.45 m/s ^2. But i don't understand why you used mgsin27. How do we know that's the force? and not mgcos27

For question 2 i don't understand how you went from force of friction = unormalforce to mgcos25 . I also don't see how it helps because we don't have the mass.
 
mgcos always equals the normal force. Unless its an object on the ground, which in that case w=n.

Alright. f=un. mgcosx25=n

f=uxmgcos25

mgxsin25=uxmgcos25

gxsin25=uxgcos25
g=9.8
solve for u
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...
Back
Top