Question

A 4-kg block is sliding down a plane as pictured, with the plane forming a 3-4-5 right triangle. Initially, the block is not moving. (However, we will assume only kinetic friction applies.) The coefficient of static friction between the block and the plane is 0.25. How long does it take for the block to reach the bottom of the plane? (Pick the answer closest to the true value.)A. 3.2 secondsB. 1.6 secondsC. 2.5 secondsD. 1.0 secondsE. 0.5 seconds

Answer 1

Answer:b

Explanation:

Given

mass of block $m=4 kg$

coefficient of static friction $\mu =0.25$

height of triangle is $h=3 m$

$F_{net}=mg\sin \theta -\mu _kmg\cos \theta$

$a_{net}=g\sin \theta -\mu _kg\cos \theta$

$a_{net}=9.8\sin 37-0.25\times 9.8\times \cos 37$

$a_{net}=5.897-1.956=3.94 m/s^2$

here $s=5 m$

$v^2-u^2=2 a_{net}s$

$v=\sqrt{2\times 3.94\times 5}$

$v=6.27 m/s$

time taken to reach bottom of plane

$v=u+at$

$6.27=0+3.94\times t$

$t=1.59 s\approx 1.6 s$