Question

An object of mass 5.0 kilograms initially travelling at12.0m/s

horizentally enters a region where there is a coefficientof Kinetic
friction of 0.27
a) Use the concepts of work and energy to determine thespeed
of the mass after it has traveled 3 meters.
b) the mass now continues its motion by sliding up a 35degree
incline.How far up the incline will it travel before comingto rest?
the coefficient of friction between the mass and theincline is
0.20.
c) Now solve the same problems in a and b usingNewton's laws-
force mass, and acceleration.

Answer 1

a) v = 11.3 m/s

b) d = 8.83 m

c) v = 11.3 m/s and d = 8.83 m

Step-by-step explanation:

a) The work-energy theorem states that the work done on an object is equal to the change in the kinetic energy of the object.

`W = K_2 - K_1\\Fx = (1)/(2)mv_2^2 - (1)/(2)mv_1^2\\-mg\mu_k x = (1)/(2)mv_2^2 - (1)/(2)mv_1^2\\-5*9.8*0.27*3 = (1)/(2)5v_2^2 - (1)/(2)5(12)^2\\v_2 = 11.3~m/s`

b) We will use the conservation of energy.

`K_2 + U_2 - W_(friction) = K_3 + U_3\\(1)/(2)mv_2^2 + 0 - mg\cos(35^\circ)\mu d = 0 + mgh\\h = d\sin(35^\circ)\\(1)/(2)5(11.3)^2 - 5(9.8)\cos(35)0.2d = 5(9.8)d\sin(35)\\319.2 - 8.02d = 28.1d\\319.2 = 36.12d\\d = 8.83 m`

c) Newton’s Second Law will be applied.

`F_(net) = ma \\-mg\mu = ma \\a = -g\mu = -2.64`

The final speed can be found by using the kinematics equations:

`v_2^2 = v_1^2 + 2ax\\v_2^2 = (12)^2 - 2(2.64)3//\\v_2 = 11.3~m/s`

Similarly,

`F_(net) = ma\\-mg\cos(35)\mu - mg\sin(35) = ma \\-g\cos(35)\mu - g\sin(35) = a\\-9.8\cos(35)0.2 - 9.8\sin(35) = a\\a =-7.22~m/s^2`

The kinematics equations are

`v_3^2 = v_2^2 + 2ad\\0 = (11.3)^2 - 2(7.22)d\\d = 8.83m`