Question

solve questions number 28​

Answer 1

♊♊YOUR ANSWERS♍♍

✍(2) 14 m/s and 15 m/s

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EXPLANATION —

↪Actually welcome to the concept of NLM and Circular Motion

↪Basically we will balance the all forces

Balancing the forces we get

Mg - N = Mv²/R

For weightlessness N = 0

Therefore

Mv²/R = Mg

Here v is speed of car and R is radius of curvature

∴v = √(Rg)

Substituting the values R = 20 m and g =10 m/s²

⇒v = √(20 × 10)

⇒v = 14.14 m/s

Therefore the speed of car at the top of the hill is between 14 m/s to 15 m/s

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Answer 2

Answer:

14.14 m/s

OPTION B :- Between 14 m/s and 15m/s

Explanation:

★ KINEMATICS AND NLM ★

★ GIVEN ;

→ WEIGHTLESSNESS AT TOP OF HILL←

» RADIUS OF CURVATURE ( R ) = 20 m

★ The speed of ROLLER COASTER at top of hill ( V )= ???

♣ WE KNOW THAT ;

★ Before Reaching at top ;

» Forces are as follow ::: ---------------

» Fg ( Gravitational ) - Fn ( Normal ) = Fc ( Centripetal )

» mg - N = mV² / R ----- ( 1 )

★ NOW ; →→→

» WHEN ROLLER COASTER REACHES AT TOP OF HILL ; IT EXPERIENCES " WEIGHTLESSNESS "

So ;

★ CONTACT FORCE / NORMAL FORCE ( N ) ≈ 0

♣ EQUATION BECOMES .....

» Mg = MV² / R ---- ( 2 )

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→ Now ; Substituting values in EQUATION " 2 " ... we get ; →→

» Mg = MV² / R

» V² = g × R

» V² = 10 × 20

★» V = √ 200 = 14.14 m/s ★«

ANSWER :- HENCE ;

FOR RADIUS OF CURVATURE OF HILL OF 20m ; THE VELOCITY OF ROLLER COASTER AT TOP OF HILL IS ;

V = 14.14 m/s .