Question

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`?

Answer 1

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`

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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