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laiz [17]
4 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]4 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]4 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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uranmaximum [27]

Hi there!

I. Only momentum is conserved.

An inelastic collision means that there is a LOSS in the KINETIC ENERGY of the system.

However, momentum is ALWAYS conserved for every type of collision unless there is an external force acting on the system.

6 0
3 years ago
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If a 4kg Bird is pushed by the window with a force of 60 N how fast is the bird accelerate?
LiRa [457]
  • Force=60N
  • Mass=4kg

\\ \ast\sf\hookrightarrow Force=Mass\times Acceleration

\\ \ast\sf\hookrightarrow Acceleration=\dfrac{Force}{Mass}

\\ \ast\sf\hookrightarrow Acceleration=\dfrac{60}{4}

\\ \ast\sf\hookrightarrow Acceleration=15m/s^2

3 0
3 years ago
an ice skater starts with a velocity 2.25 m/s in a 50.0 degree direction after 8.33s she is moving 4.65 m/s in a 120 degree dire
svetoff [14.1K]

Answer:

-0.032 m/s^2

Explanation:

To answer the question, we just need to consider the motion along the horizontal direction.

The component of the initial velocity of the ice skater along the x-direction is:

u_x = u cos \theta =(2.25)(cos 50^{\circ})=1.45 m/s

where u = 2.25 m/s is the initial velocity and 50^{\circ} is the angle.

The component of the final velocity of the ice skater along the x-direction is

v_x = u cos \theta =(4.65)(cos 120^{\circ})=-2.33 m/s

where u = 4.65 m/s is the final velocity and 120^{\circ} is the angle.

The acceleration along the x-direction is given by

a_x=\frac{v_x-u_x}{t}

where

t = 120 s is the time

Substituting,

a=\frac{-2.33-(1.45)}{120}=-0.032 m/s^2

7 0
4 years ago
If you apply force on an object such as a ball, what will possibly happen to it?​
alexandr402 [8]
It will move in the direction you pushed it, but it will go at a certain velocity
4 0
3 years ago
Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assu
baherus [9]

Answer:

823.46 kgm/s

Explanation:

At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.

So, mgh = 1/2mv²

From here, his velocity just as he reaches the surface of the water is

v = √2gh

h = 9 m and g = 9.8 m/s²

v = √(2 × 9 × 9.8) m/s

v = √176.4 m/s

v₁ = 13.28 m/s

So his velocity just as he reaches the surface of the water is 13.28 m/s.

Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.

So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.

His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,

J = (62 × 0 + 62 × 13.28) kgm/s = 0 +  823.46 kgm/s = 823.46 kgm/s

So the magnitude of the impulse J of the water on him is 823.46 kgm/s

6 0
3 years ago
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