Question

A block released at the top of a smooth inclined plane of length l, slides through distance0.25l on that plane in time T. Then time taken further by the block to reach the bottom of the plane is

Answer 1

◆【Solution】◆

Acceleration along the block = gSin(∆)

Where ∆ is the inclination of the wedge.

Acc to Newtons eq of motion

s = ut + ( 1/2 )at²

u = 0

u = 0s = l/4

l/4 = (1/2)g(Sin∆)T²

L/2 = gSin∆T²

Sin∆ = L/(2gT²)......(1)

Velocity at L/4.

v² = u² + 2gSin∆(L/4)

v² = 2g (L/2gT²)(L/4)

v² = L²/4T²

V = L/2T...(2)

Distance remaining = 3L/4

Initial velocity = L/2T

Acceleration = (L/2T²)

Time = ?

S = ut + (1/2)at²

(3L/4) = Lt/2T + (1/2)(L/2T²)(t²)

(3/4) = t/2T + t²/4T²

3T² = 2Tt + t²

t² + 2Tt - 3T² = 0

This is a quadratic equation and give two values.....

First value

t = (-2T/2) + (√4T²+12T²)/2

t = -T + 4T/2

t = T

Second value...

t = (-2T/2) - (√4T²+12T²)/2

(√4T²+12T²)/2t = -T - 4T/2

4T/2t = -T.......time never be negative

◆【Time taken = T】◆

#answerwithquality

#BAL