"on a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. wh
en max pulls on the sled with 12 n of force, directed at an angle of 15° above the ground, how much work does max do on the sled as he pulls his sister 25 m in the snow?"
1 answer:
Refer to the diagram shown below.
The angle θ = 15° relative to the direction of motion.
Therefore the force acting in the direction of motion is
(12 N) cos(15°) = 11.59 N
By definition,
Work = Force x Distance.
Therefore work done in moving the sled by 25 m is
Work = (11.59 N)*(25 m) = 289.8 J
Answer: 289.8 J
The angle θ = 15° relative to the direction of motion.
Therefore the force acting in the direction of motion is
(12 N) cos(15°) = 11.59 N
By definition,
Work = Force x Distance.
Therefore work done in moving the sled by 25 m is
Work = (11.59 N)*(25 m) = 289.8 J
Answer: 289.8 J
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Explanation:
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Rearrange to obtain an expression for the speed of light in the first medium:
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Answer:
The speed of the package of mass m right before the collision
Their common speed after the collision
Height achieved by the package of mass m when it rebounds
Explanation:
Have a look to the diagrams attached below.
a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.
where is Kinetic energy and
is Potential energy.
and
Considering the fact we will plug out he values of the given terms.
So
Keypoints:
- Sum of energies and momentum are conserved in all collisions.
- Sum of KE and PE is also known as Mechanical energy.
- Only KE is conserved for elastic collision.
- for elastic collison we have
that is co-efficient of restitution.
<u>KE = Kinetic Energy and PE = Potential Energy</u>
b.Now when the package stick together there momentum is conserved.
Using law of conservation of momentum.
where
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Plugging the values we have
Cancelling m from both sides and dividing 3 on both sides.
Law of conservation of energy will be followed over here.
c.Now the collision is perfectly elastic
We have to find the value of for m mass.
As here we can use that if both are moving in right ward with
then there is a
velocity when they have to move leftward.
The best option is to use the formulas given in third slide to calculate final velocity of object .
So
Now using law of conservation of energy.
The linear momentum is conserved before and after this perfectly elastic collision.
So for part a we have the speed for part b we have their common speed
and for part c we have the rebound height
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