Question

A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-direction is applied to the crate. The force has a constant value of 80.0 N for 12.0 s and then decreases linearly with time so it becomes zero after an additional 6.00 s. What is the final speed of the crate, 18.0 s after the force was first applied?

Answer 1

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

$F . t = m(v_f - v_i)$

so here we know that object is at rest initially so we have

$80 (12) = 80( v_f - 0)$

$v_f = 12 m/s$

Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

$F = 80 - \frac{40}{3} t$

now again by same equation we have

$\int F .dt = m(v_f - v_i)$

$\int (80 - (40/3)t) dt = 80(v_f - 12)$

$80 t - \frac{40t^2}{6} = 80(v_f - 12)$

put t = 6 s

$480 - 240 = 80(v_f - 12)$

$v_f = 12 + 3$

$v_f = 15 m/s$