A 5.20-kg package slides 1.41 m down a long ramp that is inclined at 13.0 below the horizontal. The coefficient of kinetic friction between the package and the ramp is mk = 0.313. Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.22 m>s at the top of the ramp, what is its speed after it has slid 1.41 m down the ramp?

a) Wf = -21.936 J

b) Wg = 16.18 J

c) Wn = 0 J

d) W = -5.756 J

e) v2 = 1.65 m/s

Explanation:

Given:

- The mass of the package m = 5.20 kg

- The inclination of ramp θ = 13°

- The coefficient of kinetic friction mk = 0.313

- The package slides down a distance of s = 1.41 m

- The initial velocity of the package v1 = 2.22 m/s

Find:

(a) the work done on the package by friction;

(b) the work done on the package by gravity;

(c) the work done on the package by the normal force;

(d) the total work done on the package.

(e) If the package has a speed of 2.22 m>s at the top of the ramp, what is its speed after it has slid 1.41 m down the ramp?

Solution:

- Apply Equilibrium conditions perpendicular to the surface:

                            N - m*g*cos(θ) = 0

                            N =m*g*cos(θ)

Where, N is the normal contact force exerted by the surface on the package.

- The frictional force Ff is given by:

                            Ff = mk*N

                            Ff = -mk*m*g*cos(θ)

- The work done Wf by frictional force is given by:

                            Wf = Ff*s

                            Wf = -s*mk*m*g*cos(θ)  

Plug in values:

                            Wf = -1.41*0.313*5.2*9.81*cos(13)  

                            Wf = -21.936 J

- The work done by the contact force Wn is given by:

                            Wn = s_y*N

Since, the equilibrium conditions are maintained perpendicular to surface and no distance is traveled; hence, s_y = 0

                            Wn = 0 J

- The total work done by gravity Wg is given by moving the vertical distance h from start to finish:

                            Wg = m*g*h

                            Wg = m*g*s*sin(θ)

                            Wg = 5.2*9.81*1.41*sin(13)

                            Wg = 16.18 J

- So, the total work done W is given by the sum:

                            W = Wf + Wn + Wg

                            W = -21.936 + 0 + 16.18

                            W = -5.756 J

- Construct an energy balance for the two states at top and bottom of the ramp:

                           0.5*m*v1^2 + W = 0.5*m*v2^2

                            2*(v1^2 + W/m) = v2^2

                            v2 = sqrt ( (v1^2 + 2*W/m) )

Plug in values:

                            v2 = sqrt ( (2.22^2 -5.756*2/5.2) )

                            v2 = sqrt ( 2.71455  )

                            v2 = 1.65 m/s

answer:

b) touch