Question

A sled with a mass of 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00-m rope to a post set in the ice. once given a push, the sled revolves uniformly in a circle around the post. If the sled makes five complete revolutions every minute, find the F exerted on it by the rope.​

Answer 1

Answer:

The value is  $F = 34.3 \ N$

Explanation:

From the question we are told that

    The mass of the shed is  $m = 25 \ kg$

    The length of the rope is  $l = 5.0 \ m$

    The angular speed of the shed is $w = 5 rev/minute = \frac{2* \pi * 5}{60 }= 0.524 \ rad/sec$

Generally the force exerted on the shed is mathematically represented as

     $F = m * w^2 * l$

=>  $F = 25 * 0.524^2 * 5$

=>  $F = 34.3 \ N$    

Answer 2

The magnitude of the centripetal force on the sled by the rope is 34.3 N.

The given parameters:

• mass of the sled, m = 25 kg
• length of the rope, r = 5 m
• number of revolution per minute, N = 5

The angular speed of the sled is calculated as follows;

$\omega = 5 \ \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1 \min}{60 \ s} \\\\\omega = 0.524 \ rad/s$

The magnitude of the centripetal force on the sled by the rope is calculated as follows;

$F = ma_c\\\\F = m\omega ^2 r\\\\F = 25 \times (0.524)^2 \times 5\\\\F = 34.3 \ N$

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