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Vesnalui [34]
3 years ago
8

If you throw a ball up with a velocity of 7 m/s, how long will it take for the ball to reach the top of its path?

Physics
1 answer:
nirvana33 [79]3 years ago
4 0

Answer:

c. 0.71 [s]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} =v_{o} -g*t

where:

Vf = final velocity = 0 (because when the ball reaches the top, there is no movement)

Vo = initial velocity = 7 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: The negative sign of the equation means that the movement is againts the direction of the gravity acceleration.

0 = 7 - (9.81*t)\\t = 0.713 [s]

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Answer: a) Fmax= μs.m.g  b) Ff=(μs.m.g /2  c) a= (μs.m.g - μk.m.g) / m

Explanation:

a) The friction force, is one of the components of the contact force, and can adopt any value to counteract an applied force, so the object doesn't move, till a maximum value, beyond which, if he applied force is larger, the object will start to move.

This limit value, is given by the following expression:

Fmax =  μs. N = μs. m.g

(This is valid only if the contact surface is horizontal).

 

b) As explained above, if the applied force is smaller than the limit value, the friction force will adopt the same value as the applied force, but of opposite direction, so due to the Newton's 2nd Law, the object remains at rest.

In this case, if the push is horizontal, with a force equal to the half of the limit value, friction force will be exactly equal to half the maximum friction force, as follows:

Ff = (μs. m.g) / 2

c) As the applied force is equal to the static friction force, once in movement, the opposing friction force is equal to Ffk = μk. m.g (where μk ∠ μs).

So, if the applied force is larger than this friction force (that always oppose to the relative movement between both surfaces in contact each other), the object will suffer an acceleration, which value is obtained from the Newton's 2nd Law, as follows:

Ext F = m .a ⇒ a = (μs.m.g -μk. m.g ) / m

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