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Talja [164]
3 years ago
7

A ball is rolling along at speed v without slipping on a horizontal surface when it comes to a hill that rises at a constant ang

le above the horizontal. In which case will it go higher up the hill: if the hill has enough friction to prevent slipping, or if the hill is perfectly smooth. Justify your answer with a conservation of energy statement
Physics
1 answer:
MissTica3 years ago
6 0

Answer:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

Explanation:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

This is because:

If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.

Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.

Therefore, static friction is infact necessary for a ball to begin rolling.

Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.

So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.

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A ball is dropped from a height of 20 meters. At what height does the ball have a velocity of 10 meters/second?
borishaifa [10]

Answer:B

Explanation:

Initial velocity, u=0m/s

Distance,s=20m

a=+g=9.8m/s*s

Using v*v=u*u+2gs

v*v=0+2*9.8*20

v*v=392

v=19.8

When s=20m, v = 19.8m/s

Therefore when v = 10m/s, s= 10*20/19.8

s =10.1m

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3 years ago
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tatuchka [14]

Explanation:

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8 0
3 years ago
What's the degree to which molecules are able to pass through a membrane
lawyer [7]
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3 years ago
An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm
Anit [1.1K]

Answer:

The value of d is 20.4 m.

(C) is correct option.

Explanation:

Given that,

Initial velocity = 20 m/s

Final velocity = 0

We need to calculate the time

Using equation of motion

v = u+gt

Where, u = Initial velocity

v = Final velocity

Put the value into the formula

t = \dfrac{20}{9.8}

t= 2.04\ sec

We need to calculate the distance

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

s=0+\dfrac{1}{2}\times9.8\times(2.04)^2

s=20.4\ m

Hence, The value of d is 20.4 m.

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