# IF CAN ANSWER, VERY IMPRESSIVE! A box of mass m=10kg is sliding down a ramp with angle (theta)=45 degrees.?

The coefficient of friction between the ramp and the box is U(mu)1=0.80. The box is released FROM REST in point A. In point B, the box is continuing its motion but horizontally and it sops in point C, due to friction. The coefficient of friction between the box andf the horizontal portion is U(mu)2=0.25.

(Note: The initial speed of the horizontal motion is the same with the speed at the base of the ramp.)

1. Draw free body diagram of the box on the ramp.

2. Draw free body diagram of the box on the horizontal surface.

3. Calculate the length of the ramp (A-B) if it takes the box a time t=3s to reach point B

4. Calculate the stopping distance between points B and C

Relevance

I CANNOT draw the diagram although it is the half solution. since as i saw a good diagram is giving you half the answer.

OK to work

on the ramp the Bx will push the body downwards as long as the friction is less than the Bx. (to keep it moving) so friction is

F = coeff *Nx = coef * m * g * 1/2 =0.8*10*10/2 = 40N

and Bx = 10*10/2 = 50N (since sin45=cos45=1/2)

now we have a net force of 10N to move the objet til lthe end of the ramp with an acceleration that is caused due to the net force and will end at the end of the ramo.

so

Fnet = 10N = m * a => a = 10/10 = 1m/s^2

so Sramp = 1/2 * a* t^2 = 1/2 * 1 * 9 = 4.5m

now at the end of the ramp its speed will be Urampmax= a*t = 1*3= 3m/sec.

so it starts with 1 m/se and decelerates due to new friction.

so 1/2 m u^2 = Friction * distance (conservation of energy)

so Distance (bc) = m*u^2 / 2F

f is friction = n*N= n*m*g= 10*10*0.25 = 25N ==>>

distance is 10 * 9 / 2*25 = 1.8m

please if you have the answer tell me if i am correct.