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3 years ago

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

=============

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

==========================

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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3 years ago

An object that covers more distance in the same amount of time has a higher speed.

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true

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Mode of physical activity <br> A.frequency <br> B.intensity <br> C.time <br> D.type

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A.frequency

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3 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

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(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

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$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

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$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

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3 years ago

The mass of a mass-and-spring system is displaced 10 cm from its equilibrium position and released. A frequency of 4.0 Hz is obs

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Answer:

It will remain same i.e. 4.0 Hz

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In a mass-and-spring system, the frequency depends upon the mass (m) and the spring constant (k).

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Since, the frequency does not depend on the initial displacement of the mass, the frequency would remain the same i.e. 4.0 Hz.

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