Question

A small 0.40-kg box is launched from rest by a horizontal spring as shown in the figure below. The block slides on a track down a hill and comes to rest at a distance d from the base of the hill. The coefficient of kinetic friction between the box and the track is 0.32 along the entire track. The spring has a spring constant of 37.0 N/m and is compressed 30.0 cm with the box attached. The block remains on the track at all times. Calculate d

A small 0.40-kg box is launched from rest by a horizontal spring as shown in the figure below. The block slides on a track down a hill and comes to rest at a distance d from the base of the hill. The coefficient of kinetic friction between the box and the track is 0.32 along the entire track. The spring has a spring constant of 37.0 N/m and is compressed 30.0 cm with the box attached. The block remains on the track at all times. |-30.0 cm kommer man mong 50.0 cm 40° d=24 (a) What would you include in the system? Explain your choice. In this system you would want to include all forces that are playing a roll in the boxes movement including, gravity, friction, distance, spring constant, and mass. This answer has not been graded yet. (b) Calculate d. (c) Compare your answer with a similar situation where the inclined part of the track is frictionless.

Answer 1

a) I would include frictional force which does negative work on the block.
I also include mechanical energy of the block.

b)
Workdone by friction, Wf = Workdone by friction on upper surface + on inclinded surface + on lower surface

= fk1*d1*cos(180) + fk2*d2*cos(180) + fk3*d*cos(180)

= mue_k*m*g*d1*(-1) + mue_k*m*g*cos(40)*d2*(-1) + mue_k*m*g*d*(-1)

= 0.32*0.4*9.8*0.3*(-1) + 0.32*0.4*9.8*cos(40)*0.5*(-1) + 0.32*0.4*9.8*d*(-1)

= -0.8568 - 1.2544*d

now use,

Workdone by friction = change in mechanical energy

-0.8568 - 1.2544*d = 0 -((1/2)*k*x^2 + m*g*h))

= -(1/2)*k*x^2 - m*g*d2*sin(40)

-0.8568 - 1.2544*d = -(1/2)*37*0.3^2 - 0.4*9.8*0.5*sin(40)

==> d = 1.65 m

c)

If the inclined surface is friction less,

Workdone by friction, Wf = Workdone by friction on upper surface + on lower surface

= fk1*d1*cos(180) + fk3*d*cos(180)

= mue_k*m*g*d1*(-1) + mue_k*m*g*d*(-1)

= 0.32*0.4*9.8*0.3*(-1) + 0.32*0.4*9.8*d*(-1)

= -0.3763 - 1.2544*d

now use,

Workdone by friction = change in mechanical energy

-0.3763 - 1.2544*d = 0 -((1/2)*k*x^2 + m*g*h))

= -(1/2)*k*x^2 - m*g*d2*sin(40)

-0.3763 - 1.2544*d = -(1/2)*37*0.3^2 - 0.4*9.8*0.5*sin(40)

==> d = 2.03 m