# a small block slides down on a smooth inclined plane starting from rest

### Mohammed

Guys, does anyone know the answer?

get a small block slides down on a smooth inclined plane starting from rest from screen.

## A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = ( n

Click here👆to get an answer to your question ✍️ A small block slides without friction down an inclined plane starting from rest. Let Sn be the distance travelled from time t = ( n - 1 ) to t = n . The SnSn + 1 is.

A small block slides without friction down an inclined plane starting from rest. Let SQuestion n

be the distance travelled from time t=(n−1)tot=n. The

S n+1 S n is.

**A**

2n (2n−1)

**B**

(2n−1) (2n+1)

**C**

(2n+1) (2n−1)

**D**

(2n+1) (2n) Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is C)

For distance S n

travelled from (n−1) to n

S n =0[n−n−1]+ 2 1 a[n 2 −(n−1) 2 ] S n = 2 1 a[n 2 −(n 2 +1−2n)] S n = 2 a (2n−1) Again , S n+1 = 2 1 a[(n+1) 2 −n 2 ] = 2 a [2n+1] S n+1 S n = 2 a (2n+1) 2 a (2n−1) = 2n+1 2n−1

Was this answer helpful?

483 81

## A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval, t = n 1 to t = n, then, the ratio SnSn+1 is

A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval, t = n 1 to t = n, then, the ratio SnSn+1 is

Home

A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let Sn be the distance travelled by the block in the interval, t = n 1 to t = n, then, the ratio SnSn+1 is

Question

A small block slides down on a smooth inclined plane, starting from rest at time

t = 0. Let S n

be the distance travelled by the block in the interval,

t = n − 1 to t = n , then, the ratio S n S n + 1 is A 2 n 2 n − 1 B 2 n − 1 2 n C 2 n − 1 2 n + 1 D 2 n + 1 2 n − 1 Open in App Solution

The correct option is **C**

2 n − 1 2 n + 1 Suppose, θ

is inclination of inclined plane.

So, the acceleration along inclined plane

a = g sin θ Initial speed, u = 0 .

By equation of uniformly accelerated motion, distance travelled by object during

n t h second, S n = u + a 2 ( 2 n − 1 ) S n = 0 + g sin θ 2 ( 2 n − 1 ) = g sin θ 2 ( 2 n − 1 ) . . . ( i )

Distance travelled during

( n + 1 ) t h second. S n + 1 = 0 + g sin θ 2 [ 2 ( n + 1 ) − 1 ] = g sin θ 2 ( 2 n + 1 ) . . . ( i i ) Using ( 1 ) and ( 2 ) , S n S n + 1 = ( 2 n − 1 ) ( 2 n + 1 ) Hence, ( c )

is the correct answer.

Suggest Corrections 48

SIMILAR QUESTIONS

**Q.**A small block slides down on a smooth inclined plane, starting from rest at time

t = 0. Let S n

be the distance travelled by the block in the interval,

t = n − 1 to t = n , then, the ratio S n S n + 1 is

**Q.**A block is moving down a smooth inclined plane starting from rest at time

t = 0 . Let S n

be the distance travelled by the block in the interval

t = n − 1 to t = n . The ratio S n S n + 1 is

**Q.**A smallA block slides a smooth inclined plane, starting from rest at time t = 0.Let Sn be the distance travelled by th block in the interval t=n-1 to t=n.Then the Ratioos Sn/Sn-1

**Q.**a small block slides down a smooth inclined plane,starting from rest at time t=0.let sn be the distance travelled by the block in the interval t=n-1 to t=n.Then,the ratio sn/sn+1 is: (a)2n-1/2n (b)2n-1/2n+1 (c)2n+1/2n-1 (d)2n/2n-1

**Q.**

A small block slides without friction down the inclined plane starting from rest . let Sn be the distance travelled from t = n-1 to t= n. Then Sn/Sn + 1 is ?

View More

## [Solved] A small block slides down on a smooth inclined plane, starti

CONCEPT: According to the graphical second equation of motion for an nth second we have; Sn = u + \(\frac{a}{2}\) (2n - 1) ---(1) where&n

Home Physics Motion in a Straight Line Kinematic equations for uniformly accelerated motion

## Question

Download Solution PDF

A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval

t=n−1 to t=n. Then, the ratio

n n1 SnSn+1 is : 2n2n−1 2n−12n 2n−12n+1 2n+12n−1

## Answer (Detailed Solution Below)

Option 3 : 2n−12n+1 Crack with

India's Super Teachers

FREE

Demo Classes Available*

Explore Supercoaching For FREE

**Free Tests**

View all Free tests >

FREE RPMT Optics Test 1 17 K Users

10 Questions 40 Marks 10 Mins

Start Now

## Detailed Solution

Download Solution PDF

**CONCEPT:**

According to the graphical **second equation of motion** for an nth second we have;

Sn = u + a2 (2n - 1) ---(1)

where Sn = distance traveled by an object during the nth second.

u is the initial velocity.

a is the acceleration.

**CALCULATION:**

Suppose θ is the inclination of inclined plane acceleration along the inclined plane is written as;

a = g sin θ

By using equation (1) of uniformly accelerated motion we have;

Sn = u + a2 (2n - 1)

Putting the value of acceleration 'a' and initial velocity, u is zero, we have;

Sn=0+gsinθ2(2n−1) ⇒ Sn=gsinθ2(2n−1) -----(2)

Distance traveled during (n + 1)th second.

Sn+1 = 0 + gsinθ2[2(n+1)−1] Sn+1=gsinθ2(2n−1) -----(3)

Dividing equation (2) and (3) we have;

SnSn+1=(2n−1)(2n+1)

Hence, **option 3) is the correct answer**.

Download Solution PDF

Share on Whatsapp Latest NEET Updates

Last updated on Oct 21, 2022

The National Eligibility-cum-Entrance Test (NEET) Counselling Schedule has been released for Maharashtra, Assam and Karnataka circles. Check out the detailed NEET Counselling Schedule in the linked page. The National Testing Agency (NTA) conducts NEET Exam every year for admission into Medical Colleges. The candidates can check their NEET results from the official website of NTA. For the official NEET Answer Key the candidates must go through the steps mentioned here.

Ace your Motion in a Straight Line preparations for Kinematic equations for uniformly accelerated motion with us and master Physics for your exams. Learn today!

India’s **#1 Learning** Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Get Started for Free

Download App

Trusted by 3.4 Crore+ Students

‹‹ Previous Ques Next Ques ››

## More Kinematic equations for uniformly accelerated motion Questions

**Q1.**An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is :

**Q2.**If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?

**Q3.**A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively :

**Q4.**A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is : (g = 10 m/s2)

**Q5.**A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval t=n−1 to t=n. Then, the ratio nn1

SnSn+1 is :

**Q6.**A train, which is moving at the rate of 60 miles per hour, is brought to rest in 3 minutes with a uniform retardation, find this retardation, and also the distance that the train travels before coming to rest?

**Q7.**The rear side of a truck is open and a box of 30 kg mass is placed 4.5 m away from the open end as given in the diagram. The coefficient of friction between the box and the surface below it is 0.20. On a straight road, the truck starts from rest and accelerates with 3 m/s2. At what distance from the starting point does the box fall off the truck? (ignore the size of the box) (Let g = 10 m/s2)

**Q8.**A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately

**Q9.**Which of the following equation of motion can be used to determine distance or displacement travelled by a body directly?

**Q10.**A bus at rest starts moving with an acceleration of 0.1 m/s2 . What will be its speed after 2 minutes?

## More Motion in a Straight Line Questions

**Q1.**A person travelling in a straight line moves with a constant velocity v1 for certain distance 'x' and with a constant velocity v2 for next equal distance. The average velocity v is given by the relation

**Q2.**A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time t1 if the elevator is at rest and in time t2 if the elevator is moving uniformly. Then

**Q3.**A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of the acceleration due to gravity, is _______ m. [Use g = 10 m/s2]

Guys, does anyone know the answer?