can someone help me to solve a physics problem on force and motion?

Question

this is the situation: a body of mass starts down from the top of an inclined plane 20 ft long and 10 ft high. what is its velocity at a point 12 ft from the top if the coefficient of friction is 0.1?

sorry for the pressure. thanks a lot. it will be really a big help.

Answer 1

Hint
Using Kinetic energy Ke of the body of mass m you can find the velocity V

V=sqrt(2Ke/m)

But Ke is the difference between Potential energy Pe and work done against the friction (Wf)
Ke= Pe - Wf

Pe = mgh
g= 32 ft/s^2
h= 10(12/20)=6ft
Wf= f s
f- force of friction
s- distance traveled (12ft)
f= u mg cos(A)
A=arcSin(10/20)=30 deg
the downward force is then
F= mg sin(30)

There two ways of finding Ke
1. Ke= Pe - Wf
Ke= mgh - u mg cos (30) L=0.5 mV^2
V*2=2(h - 0.5 uL )g=
V= sqrt(2 (h - u Lcos(30))g)
u=.1
h=120/20=6
L=12 feet
g= 32 ft/s^2

method 2

Ke= (F - f)L
Ke=0.5 mV^2 =[ mg sin (A) u mg cos(30)] L
then same as above

Answer 2

If theta is angle of inclination of the plane,then sin theta=10/20

sin theta=1/2,

theta =30degree

As the body starts from rest, its initial velocity u = 0

Friction force 'f'=coefficient of friction (mu) x normal reaction

f= mu x Mg cos 30

The acceleration 'a' of body of mass M=resultant force F/massM

F=[ component of weight - friction 'f' ]

F=[Mg sin 30 - mu Mg cos 30]

F =Mg [sin30 - mu cos 30]

a = F / M

a=Mg [sin30 - mu cos 30] / M

a =g [sin30 - mu cos 30]

a = g [0.5 - 0.1 x0.867]

a= g x o.4133

As g =32 ft/s^2,

a =32 x 0.4133

a =13.23 ft/s^2

If is 'v' is velocity at distance' s' =12 ft

v=sq rt 2as

v=sq rt 2 x13.23 x12

v = 17.81 ft /s

velocity at a point 12 ft from the top is 17.81 ft /s

Answer 3

If theta is attitude of inclination of the plane,then sin theta=10/20 sin theta=a million/2, theta =30degree because the body starts from leisure, its initial speed u = 0 Friction stress 'f'=coefficient of friction (mu) x common reaction f= mu x Mg cos 30 The acceleration 'a' of body of mass M=resultant stress F/massM F=[ component of weight - friction 'f' ] F=[Mg sin 30 - mu Mg cos 30] F =Mg [sin30 - mu cos 30] a = F / M a=Mg [sin30 - mu cos 30] / M a =g [sin30 - mu cos 30] a = g [0.5 - 0.a million x0.867] a= g x o.4133 As g =32 feet/s^2, a =32 x 0.4133 a =13.23 feet/s^2 If is 'v' is speed at distance' s' =12 feet v=squarert 2as v=squarert 2 x13.23 x12 v = 17.80 one feet /s speed at a component 12 feet from the optimal is 17.80 one feet /s