What is the work done by the friction when the body slides against a rough horizontal surfaces?

Physics

1 answer:

katovenus [111]2 years ago

4 0

Negative work

Hope this helps :)

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A hockey puck is sliding at a constant rate of 2m/s on a frictionless surface. How fast will the puck be moving after 10 sec

AnnyKZ [126]

2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.

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2 years ago

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$