Question

A 10 kg box initially at rest is pulled 18 meters across a horizontal, frictionless surface by a 40 N force. What is the block’s final velocity over the 18 meters?

Answer 1

As per work energy theorem we know that

Work done = change in kinetic energy

so here we will have

$W = K_f - K_i$

here we know that

work done is product of force and displacement

$W = F.d$

$W = 40(18) = 720 J$

now from above equation we have

$720 J = \frac{1}{2}(10 kg)(v_f^2 - v_i^2)$

here we know that

$v_i = 0$

so we will have

$720 = 5v^2$

$v = 12 m/s$

so its speed will be 12 m/s

Answer 2

The final velocity of the bock over the given distance is 12m/s.

Acceleration of the block

The acceleration of the block is calculated as follows;

F = ma

a = F/m

a = 40/10

a = 4 m/s²

Final velocity of the block

The final velocity of the bock is calculated using the following kinematic equation;

$v^2 = u^2 + 2as\\\\v^2 = 0 + 2(4)(18)\\\\v^2 = 144\\\\v = \sqrt{144} \\\\v =12 \ m/s$

Thus, the final velocity of the bock over the given distance is 12m/s.

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