What is the direction of the frictional force?
Answer:
Direction of friction is always opposite to relative motion
Explanation:
Friction always resist the relative motion of two surface so it is always opposite to the direction of relative velocity
What is the direction of the normal force?
Answer:
It is perpendicular to the contact plane
Explanation:
Normal force is the perpendicular force which is always at 90 degree with the contact surface
How does the frictional force depend on the normal force?
Answer:
![F_f = \mu F_n](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20F_n)
Explanation:
Friction force is directly proportional to the normal force so we can say it is given as
![F_f = \mu F_n](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20F_n)
Let the x-axis be parallel to the incline and the y-axis rise from the incline at a right angle. By Newton's second law, if there is no acceleration in the y-direction (perpendicular to the plane), what must be the magnitude of the normal force?
Answer:
![F_n = mg cos\theta](https://tex.z-dn.net/?f=F_n%20%3D%20mg%20cos%5Ctheta)
Explanation:
Normal force is equal to the component of the weight opposite to the direction of normal force
![F_n = mg cos\theta](https://tex.z-dn.net/?f=F_n%20%3D%20mg%20cos%5Ctheta)
Use Newton's second law and the x-components of the forces to find the acceleration. m/s2
Answer:
![a = gsin\theta](https://tex.z-dn.net/?f=a%20%3D%20gsin%5Ctheta)
Explanation:
As we know that
![F_{net} = ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20ma)
![mg sin\theta = ma](https://tex.z-dn.net/?f=mg%20sin%5Ctheta%20%3D%20ma)
![a = gsin\theta](https://tex.z-dn.net/?f=a%20%3D%20gsin%5Ctheta)
Answer:
Width of a box, ![l=6.41\times 10^{-15}\ m](https://tex.z-dn.net/?f=l%3D6.41%5Ctimes%2010%5E%7B-15%7D%5C%20m)
Explanation:
The ground level energy of a proton in a box is, E = 5 MeV
![E =5\times 10^6\ eV\\\\=5\times 10^6\times 1.6\times 10^{-19}\\\\=8\times 10^{-13}\ J](https://tex.z-dn.net/?f=E%20%3D5%5Ctimes%2010%5E6%5C%20eV%5C%5C%5C%5C%3D5%5Ctimes%2010%5E6%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5C%5C%5C%5C%3D8%5Ctimes%2010%5E%7B-13%7D%5C%20J)
Energy in a box is given by :
![E=\dfrac{n^2h^2}{8ml^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bn%5E2h%5E2%7D%7B8ml%5E2%7D)
For ground state, n = 1
m is mass of proton
h is Planck's constant
l is width of the box
![l^2=\dfrac{n^2h^2}{8mE}\\\\l^2=\dfrac{1^2\times (6.63\times 10^{-34})^2}{8\times 1.67\times 10^{-27}\times 8\times 10^{-13}}\\\\l=\sqrt{\dfrac{1^{2}\times(6.63\times10^{-34})^{2}}{8\times1.67\times10^{-27}\times8\times10^{-13}}}\\\\l=6.41\times 10^{-15}\ m](https://tex.z-dn.net/?f=l%5E2%3D%5Cdfrac%7Bn%5E2h%5E2%7D%7B8mE%7D%5C%5C%5C%5Cl%5E2%3D%5Cdfrac%7B1%5E2%5Ctimes%20%286.63%5Ctimes%2010%5E%7B-34%7D%29%5E2%7D%7B8%5Ctimes%201.67%5Ctimes%2010%5E%7B-27%7D%5Ctimes%208%5Ctimes%2010%5E%7B-13%7D%7D%5C%5C%5C%5Cl%3D%5Csqrt%7B%5Cdfrac%7B1%5E%7B2%7D%5Ctimes%286.63%5Ctimes10%5E%7B-34%7D%29%5E%7B2%7D%7D%7B8%5Ctimes1.67%5Ctimes10%5E%7B-27%7D%5Ctimes8%5Ctimes10%5E%7B-13%7D%7D%7D%5C%5C%5C%5Cl%3D6.41%5Ctimes%2010%5E%7B-15%7D%5C%20m)
So, the width of the bx is
.
Answer:
1176N
Explanation:
Given parameters:
Mass of bench = 100kg
Forward force = 980N
Coefficient of static friction = 0.2
Acceleration = 1.84m/s²
Unknown:
Magnitude of the applied force =?
Solution:
Frictional force is a force that opposes motion, for a body to move, the applied force must be greater than the frictional force;
The Force applied;
Force applied = Frictional force + Weight of the body
Frictional force = umg
u is the coefficient of static friction
m is the mass
g is the acceleration due to gravity
Force applied = 0.2 x 100 x 9.8 + 980 = 1176N
Given positions are
![x = 18 t](https://tex.z-dn.net/?f=x%20%3D%2018%20t)
![y = 4 t - 4.9 t^2](https://tex.z-dn.net/?f=y%20%3D%204%20t%20-%204.9%20t%5E2)
part a)
position vector is given as
![\vec r = x \hat i + y \hat j](https://tex.z-dn.net/?f=%5Cvec%20r%20%3D%20x%20%5Chat%20i%20%2B%20y%20%5Chat%20j)
![\vec r = 18 t\hat i + (4 t - 4.9 t^2) \hat j](https://tex.z-dn.net/?f=%5Cvec%20r%20%3D%2018%20t%5Chat%20i%20%2B%20%284%20t%20-%204.9%20t%5E2%29%20%5Chat%20j)
part b)
velocity is given as
![v = \frac{dr}{dt}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7Bdr%7D%7Bdt%7D)
now by differentiation of above equation
![v = 18 \hat i + (4 - 9.8 t)\hat j](https://tex.z-dn.net/?f=%20v%20%3D%2018%20%5Chat%20i%20%2B%20%284%20-%209.8%20t%29%5Chat%20j)
Part c)
For the acceleration we can use
![a = \frac{dv}{dt}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bdv%7D%7Bdt%7D)
so here it is
![a = 0 \hat i - 9.8 \hat j](https://tex.z-dn.net/?f=a%20%3D%200%20%5Chat%20i%20-%209.8%20%5Chat%20j)
Part d)
at t = 3 s
![\vec r = 54\hat i - 32.1 \hat j](https://tex.z-dn.net/?f=%5Cvec%20r%20%3D%2054%5Chat%20i%20-%2032.1%20%5Chat%20j)
![\vec v = 18 \hat i - 25.4 \hat j](https://tex.z-dn.net/?f=%5Cvec%20v%20%3D%2018%20%5Chat%20i%20-%2025.4%20%5Chat%20j)
![\vec a = - 9.8 \hat j](https://tex.z-dn.net/?f=%20%5Cvec%20a%20%3D%20-%209.8%20%5Chat%20j)