Question

A cookie jar is moving up a 25° incline. At a point 55 cm from the bottom of the incline (measured along the incline), the jar has a speed of 1.4 m/s. The coefficient of kinetic friction between jar and incline is 0.12.

(a) How much farther up the incline will the jar move? m

(b) How fast will it be going when it has slid back to the bottom of the incline? m/s

(c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)?

decrease, increase, remain the same

Answer 1

d = distance moved along the incline

V_i = initial speed = 1.4 m/s

V_f = final speed = 0 m/s

k =coefficient of kinetic friction = 0.12

from the force diagram , force equation perpendicular to incline is given as

F_n = mg Cos25

frictional force is given as

f_k = k F_n = k mg Cos25

net force parallel to incline is given as

F_net = f_k + mg Sin25 = k mg Cos25 + mg Sin25

But Fnet = ma

so ma = k mg Cos25 + mg Sin25

a = k g Cos25 + g Sin25 = 9.8 (0.12 Cos25 + Sin25)

a = 5.21 m/s²

a = - 5.21    since it is opposite to the motion

using the equation

V_f² = V_i² + 2 a d

0² = 1.4² + 2 (-5.21) d

d = 0.188 m

b)

Total distance from the bottom = L = 0.55 + d = 0.55 + 0.188 = 0.738 m

height = h = L Sin25 = 0.738 Sin25 = 0.312

using conservation of energy between Top and Bottom

Potential energy at Top = kinetic energy at bottom + work done by friction

mgh = (0.5) m V² + f_k L

mgh = (0.5) m V² + k mg Cos25 L

gh = (0.5) V² + k g Cos25 L

(9.8) (0.312) = (0.5) V² + (0.12) (9.8) (0.738) Cos25

V = 2.13 m/s

c)

Answer to a) decrease if we increase the coefficient of kinetic friction since increasing that would increase the amount of retardation and hence the jar would stop in smaller distance

Answer to b) decrease since increasing the coefficient of friction will increase the work done by frictional force and hence kinetic energy will be smaller