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postnew [5]
3 years ago
15

If the sign of work is negative,

Physics
1 answer:
aivan3 [116]3 years ago
5 0

If the sign of work is negative, that means the force and the motion are in opposite directions.

Let's say you see something roll off of the shelf. You catch it, and you let it down slowly and gently.

Gravity exerted down-force on it and it moved down. Gravity did positive work on it.

YOU exerted UP-force on it and it moved down. YOU did negative work on it.

(Also, the falling object exerted down-force on your hand, and your hand moved down. The falling object did positive work on your hand ! Where did THAT energy come from ? It came from the potential energy that the object had while it was on the shelf. Your hand absorbed that energy on the way down, doing negative work. So the object didn't have any kinetic energy when it reached the floor, and it did NOT splinter the floor or shatter in smithereens. It had barely enough energy left to make a sound when it hit the floor.)

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Explain how the extension of a spring is determined
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3 years ago
A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci
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According to the law of conservation of momentum:

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know. 

m1 = 125 kg     v1 = 12m/s      v'1 = -12.5m/s
m2 = 235kg      v2 = -13m/s     v'2 = ?

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'
(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'
1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')
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Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation. 

-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')
7.5kg.m/s=(235kg)(v_{2}')
(7.5kg.m/s)/(235kg)=(v_{2}')
0.03m/s=(v_{2}')

The velocity of the 2nd car after the collision is 0.03m/s.
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