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    Click here👆to get an answer to your question ✍️ A stone of 1 kg is thrown with a velocity of 20ms^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m . What is the force of friction between the stone and the ice?

    A stone of 1kg is thrown with a velocity of 20ms

    Question −1

    across the frozen surface of a lake and comes to rest after travelling a distance of 50m. What is the force of friction between the stone and the ice?

    Medium Open in App

    Updated on : 2022-09-05

    Solution Verified by Toppr

    Mass of the stone m=1kg.

    Initial velocity u=20ms

    −1

    Final velocity v=0 (Therefore the stone comes to rest distance travelled S=50m)

    From third equation of motion,

    v 2 =u 2 +2as (0) 2 =(20) 2 +2a(50) 100a=−400 ∴a=−4ms −2 .

    Here negative sign shows that there is relation in the motion of stone.

    Force of friction between stone and ice = Force required to stop the stone.

    =ma =1×−4=−4N or 4N.

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    A stone of 1 kg is thrown with a velocity of 20 m s^(−1) across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

    A stone of 1 kg is thrown with a velocity of 20 m s^(−1) across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

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    A stone of 1 kg is thrown with a velocity of 20 m s−1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

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    SOLUTION

    Initial velocity of the stone, u = 20 m/s

    Final velocity of the stone, v = 0 (finally the stone comes to rest)

    Distance covered by the stone, s = 50 m

    According to the third equation of motion:

    v2 = u2 + 2as Where, Acceleration, a

    (0)2 = (20)2 + 2 × a × 50

    a = −4 m/s2

    The negative sign indicates that acceleration is acting against the motion of the stone.

    Mass of the stone, m = 1 kg

    From Newton’s second law of motion:

    Force, F = Mass × Acceleration

    F = ma

    F = 1 × (− 4) = −4 N

    Hence, the force of friction between the stone and the ice is −4 N.

    Concept: Newton’s Laws of Motion - Newton's Third Law of Motion

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    A stone of 1 kg is thrown with a velocity of 20 m s^ 1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice? Physics Q&A

    A stone of 1 kg is thrown with a velocity of 20 m s^ 1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice? Check the answer now.

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    A stone of 1 kg is thrown with a velocity of 20 m s^ 1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice? Physics Q&A

    Question A stone of 1 kg

    is thrown with a velocity of

    20 ms-1

    across the frozen surface of a lake and comes to rest after travelling a distance of

    50 m

    . What is the force of friction between the stone and the ice?

    Open in App Solution

    Step 1: Given

    The initial velocity of the stone, u=

    20 ms-1

    The final velocity of the stone, v=

    0

    Distance covered by the stone, s=

    50 m Mass of stone, m = 1 kg

    Step 2: Formula used

    v2=u2+2as F=m×a

    , where

    F is force.

    Step 3: Calculation of acceleration and force of friction

    Using, v2=u2+2as 0=202+2×a×50 100a=-400 a=-400100 a=-4 ms-2

    Hence, the acceleration of stone is

    -4 ms-2

    . Here, -ve sign shows retardation.

    We know that F=m×a

    Substituting the values in the above equation

    F=m×a F=1×(-4) F=-4 N

    Here, -ve sign shows force is in the opposite direction which is the frictional force.

    Suggest Corrections 50

    SIMILAR QUESTIONS

    Q. A stone of 1 kg is thrown with a velocity of

    20 m s − 1

    across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

    Q. A stone of

    1 K g

    is thrown with a velocity of 20

    m s − 1

    across the frozen surface of a lake and comes to rest after traveling a distance of

    50 m

    . What is the force of friction between the stone and the ice?

    Q. Question 6

    A stone of 1 kg is thrown with a velocity of

    20 m s − 1

    across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

    Q.

    A stone of mass 2 k g

    is thrown with a velocity of

    20 m s − 1

    across the frozen surface of a lake and comes to rest after travelling a distance of

    100 m

    . Calculate the frictional force between the stone and the ice.

    Q.

    A stone of 1 k g

    is thrown with a velocity of

    20 m s − 1

    on the frozen surface of a lake. The stone comes to rest after travelling a distance of

    50 m

    . What is the force of friction between the stone and the ice?

    स्रोत : byjus.com

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