Question

A 5-Kg block is projected upwards with an initial speed of $10m{s}^{-1}$ from the bottom of a plane inclined at ${30}^0$ with horizontal. The coefficient of kinetic friction between the block and the plane is 0.2.
a. How far does move up the plane?
b. How long does it move up plane?
c. After what time from its projection does the block again come back to the bottom? With what speed does it arrive?

Answer 1

$5gsin{30}^0$ and ${\mu }_{K}N$ both act downwards, so acceleration a will be downward. Therefore,
$\therefore a=\frac{5gsin{30}^0+{\mu }_{k}N}{5}$
= $gsin{30}^0+{\mu }_{k}gcos{30}^0$
= $6.7m{s}^{-2}$ $(\because N=5gcos{30}^0)$
a. Final velocity is zero. Let the block move up the distance l before stopping: $0^2=u^2-2al$
$\Rightarrow l=\frac{u^2}{2a}=\frac{{10}^2}{2\times 6.7}=7.46m$
b. $v=u+at\Rightarrow 0=10-6.7t\Rightarrow t=10/6.7=7.46=1.5s$
c. In this case, friction will act in upward direction
$a=\frac{5gsin{30}^0-{\mu }_{k}N}{5}$
$=3.26m{s}^{-2}$
Let the time taken be t. Block will start from rest,
$l=\frac{1}{2}a{t}^2\Rightarrow t=\sqrt{2l/a}=\sqrt{\frac{2\times 7.46}{3.26}}=2.14s$
Total time = 1.5 + 2.14 = 3.64 s