Answer:
The block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8.
Explanation:
Given:
Mass of the block (m) = 3.6 kg
Initial velocity (u) = 1.7 m/s
Final velocity (v) = 0 m/s
Displacement (S) = 1.6 m
First we will find the acceleration of the block.
Using the equation of motion, we have:
[tex]v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}[/tex]
Now, plug in the given values and solve for 'a'. This gives,
[tex]a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2[/tex]
The acceleration is negative as it is resisting the motion.
Now, the initial velocity is increased by a factor of 2.8. So,
New initial velocity = 2.8 × 1.7 = 4.76 m/s
Again using the same equation of motion and expressing the result in terms of 'S'. This gives,
[tex]v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}[/tex]
Now, plug in the given values and solve for 'S'. This gives,
[tex]S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m[/tex]
Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8
