Question

A block of mass 3.3 kg, sliding on a horizontal

plane, is released with a velocity of 1.8 m/s.
The blocks slides and stops at a distance of
1.5 m beyond the point where it was released.
How far would the block have slid if its
initial velocity were quadrupled?

Answer 1

24 m

Step-by-step explanation:

initial velocity is u = 3.5 m/sec

final velocity is v = 0 m/sec

distance traveled is S = 1.5 m

From kinematic equation
`v^(2)-u^(2) = 2as`

Making acceleration the subject,

`a=\frac {v^(2)-u^(2)}{2s}`

Acceleration
`a = \frac {0^2-3.5^2}{(2*1.5)}= -4.083333333m/s^(2)`

When the initial velocity is quadrupled= u = 3.5*4 = 14 m/s

V = 0 m/s

`a = -4.083333333m/s^(2)`

then
`S = \frac {v^(2)-u^(2)}{2} = \frac {0^(2)-14^(2)}{2* -4.083333333} = 24 m`