Respuesta :
ANSWER
0.51 m
EXPLANATION
Given:
• The mass of the block, m = 0.900 kg
,• The block is initially at rest, so its initial velocity u = 0 m/s
,• There is no friction
,• The force applied horizontally, F = 8.50 N
,• The distance the force is applied for, d = 4.00 m
Find:
• How much farther would the force have to act, x, for the block to have a speed of 11.8 m/s?
The work done by the force applied to the block, W, is equal to the change in its kinetic energy, KE,
[tex]KE=W=F\cdot d=8.50N\cdot4.00m=34J[/tex]We know that the block was at rest, so the change in its kinetic energy is,
[tex]KE=\frac{1}{2}m(v-u)^2=\frac{1}{2}mv^2[/tex]We can find what is the speed of the block after the first 4 meters,
[tex]v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\cdot34J}{0.9kg}}\approx8.69m/s[/tex]Now, we know that the force must act for a little farther so that the block has a speed of 11.8 m/s. In that distance, the block's kinetic energy changes as,
[tex]KE_f=\frac{1}{2}m(v_f-v)^2=\frac{1}{2}\cdot0.9kg\cdot(11.8m/s-8.69m/s)^2\approx4.35J[/tex]Following the same reasoning as before, this change in kinetic energy must be made by the work done, so, if the force applied is the same,
[tex]W_f=KE_f=F\cdot x[/tex]Solving for x,
[tex]x=\frac{W_f}{F}=\frac{KE_f}{F}=\frac{4.35J}{8.50N}\approx0.51m[/tex]Hence, the force must be applied for another 0.51 m, rounded to two decimal places.
