Question

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.50 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 6.00x10^-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^2Hint 1. How to approach the problem Hint 2. Find the total work done on the box Hint 3. Initial kinetic energy Hint 4. What is the final kinetic energy?

Answer 1

Wtotal = KEfinal - KEinitial

42.81 J + 1.47 N * u^2 * 1.70 m/kg

Step-by-step explanation:

To solve this problem, we can use the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy. We can break the problem down into several steps to determine the total work done on the box and its final kinetic energy.

Step 1: Find the work done by gravity

The gravitational force acting on the box as it moves up the incline is given by Fg = mgsin(theta), where m is the mass of the box, g is the acceleration due to gravity, and theta is the angle of the incline. The work done by gravity over a vertical distance h is given by Wg = Fg*h. Plugging in the values given in the problem, we get:

Fg = 2.50 kg * 9.81 m/s^2 * sin(30.0°) = 12.23 N

Wg = Fg * 3.50 m = 42.81 J

Step 2: Find the work done by friction

The frictional force acting on the box is given by Ff = mgcos(theta)mu, where mu is the kinetic coefficient of friction. The work done by friction over a horizontal distance s is given by Wf = Ffs. Since the box starts from rest and comes to a stop, the work done by friction must be equal to the initial kinetic energy of the box. Plugging in the values given in the problem, we get:

Ff = 2.50 kg * 9.81 m/s^2 * cos(30.0°) * 6.00x10^-2 = 1.47 N

Wf = Ff * s

To find the distance s, we can use the fact that the box comes to a stop at the highest point of its trajectory, where its final velocity is zero. Using the equations of motion, we can find the distance s:

v^2 - u^2 = 2as

where v is the final velocity (0 m/s), u is the initial velocity, a is the acceleration (given by Ff/m), and s is the distance traveled.

Plugging in the values given in the problem, we get:

0 - u^2 = 2 * (1.47 N) / 2.50 kg * s

s = u^2 / (2 * (1.47 N) / 2.50 kg) = u^2 * 1.70 m/kg

Step 3: Find the total work done on the box

The total work done on the box is the sum of the work done by gravity and the work done by friction:

Wtotal = Wg + Wf = 42.81 J + Ff * s

Plugging in the values of Ff and s, we get:

Wtotal = 42.81 J + 1.47 N * u^2 * 1.70 m/kg

Step 4: Find the initial kinetic energy and final kinetic energy

The initial kinetic energy of the box is given by KE = 1/2 * m * u^2, where m is the mass of the box and u is the initial velocity. The final kinetic energy of the box is zero, since it comes to a stop at the highest point of its trajectory.

Therefore, the initial kinetic energy is:

KE = 1/2 * 2.50 kg * u^2

And the final kinetic energy is:

KE = 0 J

Putting it all together, we get:

Wtotal = KEfinal - KEinitial

42.81 J + 1.47 N * u^2 * 1.70 m/kg