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BigorU [14]
3 years ago
7

A block of mass 3.6 kg, sliding on a horizontal plane, is released with a velocity of 1.7 m/s. The block slides and stops at a d

istance of 1.6 m beyond the point where it was released. How far would the block have slid if its initial velocity were increased by a factor of 2.8
Physics
1 answer:
pentagon [3]3 years ago
6 0

Answer:

The block would have slid <u>12.6 m</u> 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:

v^2=u^2+2aS\\\\a=\dfrac{v^2-u^2}{2S}

Now, plug in the given values and solve for 'a'. This gives,

a=\frac{0-1.7^2}{2\times 1.6}\\\\a=\frac{-2.89}{3.2}=0.903\ m/s^2

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,

v^2=u^2+2aS\\\\S=\dfrac{v^2-u^2}{2a}

Now, plug in the given values and solve for 'S'. This gives,

S=\frac{0-4.76^2}{2\times -0.903}\\\\S=\frac{-22.6576}{-1.806}=12.6\ m

Therefore, the block would have slid 12.6 m if its initial velocity were increased by a factor of 2.8

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The momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

<h3> MOMENTUM:</h3>

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