Please show the equation and process to solve it and do it in a way that you wou

Question

Please show the equation and process to solve it and do it in a way that you would show someone who has no idea what your talking about. Thanks!

Using a simply pulley/rope system, a crewman on an Arctic expedition is trying to lower a 6.95-kg crate to the 25.9 meters. The 60.9-kg crewman is walking along holding the rope, being careful to lower the crate at a constant speed of 1.50 m/s Unfortunately, when the crate reaches a point 14.4 meters above the ground, the crewman steps on a slick patch of ice and slips. The crate immediately accelerates toward the ground dragging the hapless crewman across the ice and toward the edge of the cliff 60.9 kg of a steep ravine of height frictionless 6.95 kg 25.9 m 14.4 m If we assume the ice is perfectly slick (that is, no friction between the crewman and the ice once he slips and falls down), at what speed will the crate hit the ground? Assume also that the rope is long enough to allow the crate to hit the ground before the crewman slides over the side of the cliff figure not to scale Number At what speed will the crewman hit the bottom of the ravine? (Assume no air friction.) Number m s

Explanation / Answer

The very first thing we need to do is find the crate's acceleration as it falls. It will be less than 9.8 m/s² (the standard acceleration due to gravity for objects near Earth's surface) because the crewman's mass will provide some extra inertia. We'll set up net force equations for both the crewman and the crate.

For the crewman, the net force amounts to the force of the rope pulling him horizontally along the ice.
Fnet = ma = Frope
(60.9 kg)a = Frope

For the crate, the net force is the difference between the rope's force pulling upward and gravity pulling downward. Since the crate is accelerating downward, gravity must be the stronger force:
Fnet = Fw - Frope
ma = mg - Frope
(6.95 kg)a = (6.95 kg)(9.81 m/s²) - Frope
(6.95 kg)a = 68.11 N - Frope

Pulling these two equations together gives us:
(60.9 kg)a = Frope
(6.95 kg)a = 68.11 N - Frope

Adding them together....
(67.85 kg)a = 68.11 N ======> a = 1.004 m/s^2

Since they're connected by a rope (that presumably has no mass and doesn't stretch), we may assume that the crate's downward acceleration is equal to the crewman's horizontal acceleration.

Now we can work out the crate's impact velocity.

v(i) = 1.50 m/s
a = 1.004 m/s²
d = 14.4 m
v(f) = ???

v(f)² = v(i)² + 2ad
v(f)² = (1.50 m/s)² + 2(1.004 m/s²)(14.4 m)
v(f)² = 2.25 m²/s² + 28.92 m²/s²
v(f)² = 31.17 m²/s²
v(f) = 5.58 m/s

b) The crewman's case is slightly more complicated; once he leaves the edge of the ravine he becomes a projectile. We need to work out his horizontal velocity, then determine his vertical velocity at the moment of impact, then use the Pythagorean theorem to calculate his actual speed.

The crewman is accelerated by the rope for as long as the crate is falling, which means that his horizontal velocity will reach 5.58 m/s. Since he has no friction with the ice, he won't be able to slow down before sliding off the edge of the ravine. So his initial horizontal velocity is 5.58 m/s.

His initial vertical velocity is 0 m/s, but that will change quickly. Since he's not supported by a rope, he WILL accelerate downward at 9.81 m/s², over a distance of 25.9 meters (the full height of the ravine).

v(f)² = v(i)² + 2ad
v(f)² = (0 m/s)² + 2(9.81 m/s²)(25.9 m)
v(f)² = 508.158 m²/s²
v(f) = 22.54 m/s

Now we have the crewman's horizontal and vertical velocities. To find his speed, use the Pythagorean theorem.

s² = (5.58 m/s)² + (22.54 m/s)²
s² = 31.14 m²/s² + 508.05 m²/s²
s² = 539.19 m²/s²
s = 23.22 m/s

I hope that helps. Good luck!

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---