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 2.90 kg, up an incline of constant slope angle 30.0degree so that it reaches a stranded skier who is a vertical distance 2.60 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00 times 10^-2. 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/s^2. 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.

Explanation / Answer

The force of kinetic friction is

F_kf = F_normal = m g cos().

The length, s, of the slope is:

s = h / sin()

So the work against the friction will be:

W = F s = m g cos() * h / sin()
= m g h / tan() .

Also, the box will have to be brought to a higher gravitational potential energy, which is
U = m g h. (This is the same as saying it has to perform work: a force m g 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 m v^2 = m g h / tan() + m g h

v^2 = 2 g h / tan() + 2 g h = 2 g h ( 1 + / tan() )

v = ( 2 g h ( 1 + / tan() ) )
= ( 2 * 9.81 m/s^2 * 2.60 m * ( 1 + 6.00*10^-2 /tan(30.0)) )
= 7.36 m/s

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