Question

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 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 6.00×10⁻². 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/s2.

Part A. 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.

Answer 1

Using Energy conservation at the bottom and the point where skier is stranded

KEi + PEi + Wf = KEf + PEf

KEf = 0, since final speed of box is zero

PEi = 0, since at bottom of incline h is zero

KEi = (1/2)*m*V^2

PEf = m*g*h

m = mass of box = 2.20 kg

V = minimum speed required for box to reach skier = ?

Now there is work-done only due to friction force, So

Wf = Ff*d*cos

Ff = Force of friction = *N = *m*g*cos

d = displacement of box along the incline = h/sin

h = vertical height of incline = 3.50 m

So,

Wf = (*m*g*cos )*(h/sin )*cos

= angle between friction force and displacement = 180 deg, since friction force is always in opposite direction of motion

= incline angle = 30 deg

= coefficient of kinetic friction = 6.00*10^-2

So,

KEi + PEi + Wf = KEf + PEf

(1/2)*m*V^2 + 0 + (*m*g*cos )*(h/sin )*cos = 0 + m*g*h

divide by 'm'

(1/2)*V^2 + (*g*cos )*(h/sin )*cos = g*h

Using given values:

(1/2)*V^2 + 0.06*9.81*3.50*cos 30 deg*cos 180 deg/sin 30 deg = 9.81*3.50

(1/2)*V^2 - 3.568 = 34.335

V = sqrt (2*(34.335 + 3.568))

V = 8.71 m/sec = minimum speed required

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