How much work does it take to push an object with a mass of #2 kg# up a #2 m# ramp, if the ramp has an incline of #(11pi)/12 # and a kinetic friction coefficient of #3 #?

1 Answer
Nov 21, 2017

My calculator is in the degree mode, so I will convert #(11pi)/12 "to" ^@#.

#(11cancel(pi))/12 * 180^@/cancel(pi) = 165^@#

Note: #pi/2 = 90^@# so this ramp is inclined so far that it is almost upside down. Another #15^@# and it would be level.

I prefer to use conservation of energy. The top of the ramp has gravitational potential energy of
#m*g*h = 2 kg*9.8 m/s^2*2 m*sin165 = 10.1 kg*m^2/s^2 = 10.1 J#

In addition to that, work was done against the force of friction -- call it #F_f#. That work is
#"work" = F_f*s = mu_k*N*s = 3*m*g*cos165*2 m#
# "work" = 114" "kg*m^2/s^2 = 114 J#

So the sum is #10.1 J + 114 J ~= 124 J#


Now I will calculate work using the force *distance method. (Because cos165 has a negative value, I will use cos15 instead of cos165 in the following steps.)
The force of friction from the work above is
#F_f = 3*m*g*cos15 = 3*2 kg*9.8 m/s^2*cos15 #
#F_f = 56.8 kg*m/s^2 = 56.8 N#

Exerting that force over the 2 m length calculates to #2 m*56.8 N = 114 J#
There is also some work against gravity. The downslope component of the weight of that mass, #W_(ds)# is
#W_(ds) = m*g*sin165 = 2 kg*9.8 m/s^2*sin165 = 5.07 N#

So the work of pushing against that force is #5.07 N*2 m = 10.14J#

So the sum is #10.1 J + 114 J ~= 124 J#

I hope this helps,
Steve