You are a member of an alpine rescue team and must get a box of supplies, with m

Question

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.00 kg , up an incline of constant slope angle 30.0 so that it reaches a stranded skier who is a vertical distance 3.50 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00×102. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s2 .

Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. Express your answer numerically, in meters per second.

Explanation / Answer

The force of kinetic friction is

F_f = u*F_normal = u*mgcos(theta)

The length, s, of the slope is

s = h / sin(theta)

So the work against the friction will be:

W = F_f*s = u*mgcos(theta) * h / sin(theta)

W = u*mgh / tan(theta) .

Also, the box will have to be brought to a higher gravitational potential energy, which is

U = mgh (This is the same as saying it has to perform work: a force mg over a distance h)

So the kinetic energy at the bottom of the slope must equal the work to be done against friction plus the potential energy gain

(1/2)mv^2 = u*mgh / tan(theta) + mgh

v^2 = 2u*gh / tan(theta) + 2gh = 2gh ( 1 + u / tan(theta) )

v = sqrt ( 2gh ( 1 + u / tan(theta) ) )

v = sqrt ( 2 * 9.81 * 3.50 * ( 1 + 0.06 / tan(30)) )

v = 8.7 m/s

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