# How Quickly Can a Rock Be Hoisted with Limited String Strength?

• 8parks11
In summary, the minimum time to hoist a 1.00-kg rock with a breaking strength of 10.8 N and a height of 10.0m is 4.5 seconds. By replacing the string with one that is 50% stronger, the minimum time for the hoist will be reduced by a certain percentage that can be solved by using the information provided in part a) and using the hints provided.
8parks11
a) What is the minimum time in which one can hoist a 1.00-kg rock height of 10.0m if the string used to pull the rock up has a breaking strength of 10.8 N? Assume the rock to be initially at rest.

b) If the string is replaced by one that is 50% stronger, by what percentage will the minimum time for the hoist be reduced.

a)

i just used a= 2(x-xo)/t^2 and then i got 2(10)/t^2 = 10.8/1
so i got 10.8=20/t^2 and so t would be root 10.8 but the answer is 4.5seconds... how?

b) i need to get a) to solve this...

Hint 1: Draw a free body diagram for the rock.
Hint 2: There is some provided info in the problem you haven't used yet.

I would approach this problem by first analyzing the given information and setting up equations to solve for the minimum time needed to hoist the 1.00-kg rock.

Firstly, we know that the maximum tension the string can withstand is 10.8 N. We also know that the rock is initially at rest, so its initial velocity is 0 m/s. We can use the equation F=ma to calculate the minimum force needed to hoist the rock, which is equal to its weight, mg. Therefore, we have:

F = mg = (1.00 kg)(9.8 m/s^2) = 9.8 N

Since the maximum tension the string can withstand is 10.8 N, we can conclude that the minimum force needed to hoist the rock is 9.8 N. Now, we can use the equation F=ma again, but this time we will use the minimum force needed (9.8 N) as the force and the acceleration as the unknown variable. This gives us:

9.8 N = (1.00 kg)a

Solving for a, we get a = 9.8 m/s^2.

Next, we can use the kinematic equation x = xo + vot + 1/2at^2 to calculate the minimum time needed to hoist the rock. Since the rock starts from rest (vo = 0 m/s) and reaches a height of 10.0 m (x = 10.0 m), we have:

10.0 m = 0 + 0t + 1/2(9.8 m/s^2)t^2

Solving for t, we get t = 2.02 seconds. This is the minimum time needed to hoist the rock to a height of 10.0 m with a string breaking strength of 10.8 N.

Moving on to part b, if the string is replaced with one that is 50% stronger, its breaking strength will be 10.8 N x 1.5 = 16.2 N. Using the same approach, we can calculate the minimum force needed to hoist the rock with this stronger string, which is 9.8 N. However, since the force is now greater, the acceleration will also be greater. Using the same equation as before, we get:

9.8 N = (1

## What is the concept of "Minimum time for a tension"?

The concept of "Minimum time for a tension" refers to the amount of time required for a system to reach a state of equilibrium or balance when subjected to an external force or tension.

## Why is the minimum time for tension important?

The minimum time for tension is important because it helps us understand the behavior of systems under external forces and allows us to predict how long it will take for the system to reach a state of equilibrium.

## How is the minimum time for tension calculated?

The minimum time for tension is calculated by using mathematical equations that take into account factors such as the magnitude of the external force, the properties of the system, and any other relevant variables.

## Can the minimum time for tension be shortened?

Yes, the minimum time for tension can be shortened by increasing the magnitude of the external force or by changing the properties of the system in a way that reduces the time needed to reach equilibrium.

## What are some real-life applications of the concept of minimum time for tension?

The concept of minimum time for tension has many practical applications, such as in engineering, where it is used to design structures that can withstand external forces, and in physics, where it is used to study the behavior of systems under different conditions.

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