Engineering
Physics
Friction
Question

A shot putter with a mass of 80 kg pushes the iron ball of mass of 6 kg from a standing position, accelerating it uniformly from rest at an angle of 45° with the horizontal during a time interval of 0.1 seconds. The ball leaves his hand when it is 2 m high above the level ground and hits the ground 2 seconds later.

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Linked Question 1

The minimum value of the static coefficient of friction if the shot putter does not slip during the shot is closest to :

0.48

0.28

0.38

0.58

Solution
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Fy – 60 = 90 × 6    ⇒    Fy = 540 + 60 = 600 N

Fx = 6 × 90 = 540 N
N = Fy + Mg = 600 + 800 = 1400 N                


mN = Fx     ⇒    m = 540/1400 = 0.38

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Linked Question 2

The acceleration of the ball in shot putter's hand

902m/s2

1002m/s2

92m/s2

112m/s2

Solution
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h=uyt+12ayt2

2=u2212×10×4

u=92m/s

a=920.1=902m/s2

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Linked Question 3

The horizontal distance between the point of release and the point where the ball hits the ground

18 m

22 m

16 m 

20 m

Solution
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x=u2t=18m

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