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laiz [17]
3 years ago
8

After a short time, the moving sled with the child aboard reached a rough level surface that exerts a constant frictional force

of 54 Newtons on the sled. How much work must be done by friction to bring the sled with the child to a stop

Physics
2 answers:
valentina_108 [34]3 years ago
5 0

Answer:

Wf = –750J

Explanation:

The full solution can be found in the attachment below.

This problem involves the concepts of conservation of momentum and energy.

The potential energy of the sled and child combined is constant since thehy are moving in a level surface.

To find the initial velocity of the sled with child we use the principle of momentum conservation to calculate the v1 which is the commovelocity of the child and sled.

Alex73 [517]3 years ago
5 0

Answer:

Incomplete question

Complete question

A 50.-kilogram child running at 6.0 meters per second jumps onto a stationary 10.-kilogram sled. The sled is on a level frictionless surface.

After a short time, the moving sled with the child aboard reached a rough level surface that exerts a constant frictional force of 54 Newtons on the sled. How much work must be done by friction to bring the sled with the child to a stop.

Explanation:

The mass of the boy is

Mb=50kg

Mass of the sled

Ma=10kg

Velocity of the boy

Speed vb=6m/s

Now, we need to know the velocity of the boy and the sled Vt

Using law of momentum

Momentum before collision = momentum after collision

Mb×Vb=(Mb+Ms)Vt

50×6=(50+10)Vt

300=60Vt

Vt=300/60

Vt=5m/s

The kinetic energy of both the boy and the sled is the workdone by friction to to stop the sled

Kinetic enemy is given as

K.E=½mv²

K.E=½×60×5²

K.E=30×25

K.E=750J

So the work done by friction to stop the boy and the sled is -750J

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Answer:

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Explanation:

Since each skater pushes the other with a force of 70 N, according to Newton's third law, there is an equal reaction and thus the other pushes back with a force of 70 N in the other direction, so we have forces of +70N and -70 N respectively. So, the net force on each skateboarder is F = + 70 N + (-70 N) = + 70 N - 70 N = 0 N.

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