Question

block of mass 2.25 kg starts out at the rest at the top of 15.0 m long plane inclined 30° with horizontal .a)what is its initial gravitational potential energy.? as.it slides down the plane,the block encounters.frictional force at 3.5n. b)how much work is done to overcome.friction. c)what is the kinetic energy of the block when it reaches the bottom od the incline?at a distance of .5m from the bottom of the inclined plane is horizontal.spring of the force constant 52n/m. suppose that the coefficient of kinetic friction between the block and the horizontal surface is 0.15? d)what is the kinetic energy of the block just before it comes in contact with the spring? e)by how much will the springbe compressed? f)what is the work done against friction as the block moves along the horizontal surface? is

Answer 1

$m = 2.5kg$

$l= 15m$

$g\approx 10\frac{m}{s^2}$

$\angle\alpha = 30\degree$

$a)$

$EP = mgh$

$h = \sin\alpha*l=7.5 m$

$EP = 2.5*10*7.5 = 187.5J$

$b) \newline F_{fr}= 3.5N$

$W_{fr}= F_{fr}l=3.5*15= 52.5J$

$c)$

$EK= EP -W{fr}= 187.5-52.5=135J$

$d)$

$\mu_1= 0.15$

$s_1=0.5m$

$EK= 135J$

$EK_1= EK-W{fr_1}$

$F_{fr_1}= \mu_1mg = 0.15*2.5*10= 3.75N$

$W_{fr_1}= F_{fr_1}s_1=3.75*0.5\approx0.56J$

$EK_1=135-0.56=134.44J$

$e)$

$k=52\frac{N}{m}$

$EK_1= 134.4J$

$EP_{sp}= \frac{kx^2}{2}$

$EK_1-W_{fr_2}=EP_{sp}$

$W_{fr_2}=F_{fr_1}*x= 3.75x$

$134.4-3.75x = 26x^2;x>0$

$x=2.21m$

$f)$

$W_{fr_2}=F_{fr_1}*(s_1+x)$

$W_{fr_2}= 3.75*2.71=10.16J$

$\text{Answer:}$

$a)187.5J$

$b)52.5J$

$c)135J$

$d)134.44J$

$e)2.21m$

$f)10.16J$