A man carries a load of 20N on his head over a horizontal distance of 20 m how much work is done by the man

1 answer:

4 0

Work done in horizontal direction = 0 as the potential energy is not changed and the change in kinetic energies : initial and final - is zero.

work done when the person carries the suitcase in vertical direction =

= m g h = change in potential energy = 30 kg * 9.8 m/sec/sec * 10 m

= 2940 Joules

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There is a force F1 lifting the body upwards. There is Weight W = m g pulling the body downwards. The net force F2 on the body is upwards with acceleration a.

F2 = m a = F1 - W = F1 - m g

a = F1 / m - g

Work done by Force F1 with which the object is lifted = F1 . s = 230 Joules

F1 = energy spent / distance = 230 J / 2 m = 115 Newtons

Net Acceleration of the body = a = F1 / m - g = 115/10 - 10 = 1.5 m/sec/sec

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m = 2 kg F = 8 Newtons t = 12 sec

a = F/m = 4 m/sec/sec

v = u + a t = 0 + 4 * 12 = 48 m/sec

kinetic energy = 1/2 m v² = 1/2 * 2 * 48 * 48 = 48² Joules

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true for first and false for second

Explanation:

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The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g

Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

$k=\sqrt{\frac{I}{M} }$

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration $k=\sqrt{\frac{I}{M} }$

So, moment of rotational inertia (I) of a cylinder about it axis = $\frac{MR^2}{2}$