Question about frictionless motion with variable acceleration.

In summary, the problem involves a person climbing a mountain with a rope attached to a rock on a frictionless surface. The person (100 kg) is 3 meters from the flat surface, while the rock (700 kg) is 6 meters from the edge. The minimum acceleration the person needs to climb up at to reach the edge before the rock does can be found by writing equations for the tension (T) and normal force (N) in both the x and y directions. The person's upward acceleration must be double the downward acceleration, and the time it takes to reach the top can be equated to the time it takes for the rock to reach the edge using the formula s = 0.5a*t^2.
  • #1
thenamestiki
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1. Person is climbing a mountain with rope attached to rock on a flat surface above for support. The slope has an incline angle of 50 degrees. The rock is on frictionless ground, and the mass of the rope can be ignored. The person (100 kg) is 3 meters from the flat surface, the rock (700 kg) is 6 meters from the edge. What is the minimum acceleration the person needs to climb up at to reach the edge before the rock does. All surfaces are frictionless.



2. I do not know what equations to use :(



3. I attempted by finding the downward acceleration of the person, when he is not moving, and finding that his upward acceleration must be double that.
 
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  • #2
How about writing the equations relating T, the tension in the rope, W, the weight of the man = mg, and N, the force normal to the slope, in both x and y directions?
2 equations, 2 unknowns (T and N).

Remember that T must include the force supporting his upward acceleration
s_double-dot along the slope , where ds = d(x_m)*sec(theta) and theta = 50 deg.
(He starts at s = 0 and winds up hopefully at s = +3m). x_m is the x component of s.

Then, write the corresponding equation for the rock's force and acceleration, involving T, M and x_M where M = mass of rock and x_M is the (negative) distance the rock moves. x_M = 0 for the rock at start of slide and = -6m at time m reaches the top and the rock reaches the edge.

PS - how did you determine that

Then equate time t(s = +3m) = t(x_M = -6m) using the s = 0.5a*t^2 formula.

ps how did you "... attempt by finding the downward acceleration of the person, when he is not moving, and finding that his upward acceleration must be double that." without knowing what equations to use?
 

Related to Question about frictionless motion with variable acceleration.

1. What is frictionless motion with variable acceleration?

Frictionless motion with variable acceleration refers to an object moving without any resistance or friction, while also experiencing a change in its acceleration over time.

2. How is frictionless motion with variable acceleration different from regular motion?

Regular motion involves objects experiencing friction, which can slow down or hinder their movement. Frictionless motion eliminates this resistance, allowing for a smoother and more constant acceleration.

3. What are some examples of frictionless motion with variable acceleration?

Some examples include a car moving on an icy road, a ball rolling on a perfectly smooth surface, or a pendulum swinging in a vacuum.

4. What factors affect frictionless motion with variable acceleration?

The main factor that affects frictionless motion with variable acceleration is the object's mass. The greater the mass, the more force is needed to accelerate it. Additionally, any external forces, such as gravity or air resistance, can also impact the object's motion.

5. How is frictionless motion with variable acceleration used in science and engineering?

Frictionless motion with variable acceleration is often used in theoretical physics and engineering to model and predict the behavior of objects in ideal conditions. It can also be used in experiments to study the effects of friction and other external forces on an object's motion.

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