# a stone of 1 kg is thrown with a velocity of 20m/s across the frozen

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

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 20msQuestion −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?

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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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Chapter 9 Force and Laws 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.

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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?

Guys, does anyone know the answer?