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3 years ago

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Two objects are moving at equal speed along a level, frictionless surface. the second object has twice the mass of the first obj

ect. they both slide up the same frictionless inclined plane. which object rises to a greater height?

1 answer:

Zarrin [17]3 years ago

5 0

first object gors highest

second object goes lowest.

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4 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