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2 years ago

An elevator is moving down with an acceleration of 3.36 m/s2.

Physics

1 answer:

sergeinik [125]2 years ago

5 0

Answer : 413.44N

Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .

From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
When a elevator accelerates down , the weight recorded is less than the actual weight .

From the Free body diagram ,

$\sf\longrightarrow Weight = mg - ma \\$

$\sf\longrightarrow Weight = m ( g - a ) \\$

Mass of the man = 64.2 kg

$\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\$

$\sf\longrightarrow Weight = 64.2 * 6.44 N\\$

$\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\$

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8 0

3 years ago

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago

Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 21.0 kg and box B has mass 8.0 kg. A horizont

vfiekz [6]

Answer:

2.75 N

Explanation:

Given that,

Box A has a mass 21.0 kg and box B has a mass 8.0 kg.

A horizontal force of 100N is exerted on box A.

Let a be the acceleration of the system. Using second law of motion,

$F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2$

Now applying Newton's second law to box B. So,

$F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N$

So, 2.75 N is the force that box A exerts on box B.

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