Question

a block of mass m=10kg is released from rest on a frictionless incline of angle=30°. Theass can be compressed 2.0cn by a force of 200N. The block momentarily stops when it compresses the spring by 5cm.. How far does fge block mobe down fge incline from its rest position to its stopping point. and what is the speedbof the block just as it touches the sprinv

Answer 1

the energy at max compression is 1/2 k x^2=1/2 *200N/.02m * .05^2=12.5N
Height it goes upward:
12.5=mgh=10*9.8*h or h=12.5/98 m=12.75cm
which is a distance up the ramp of 12.75/sin30=25.5cm
speed of block as it touches spsring; 1/2 mv^2=12.5 solve for v

Answer 2

Gravitational force downhill = m g sin 30 = 98/2 = 49 Newtons
moves down slope distance z
compresses spring distance x = 0.05
energy stored in spring at stop = (1/2)(200/0.02)(0.05)^2 = 12.5 Joules
so loss in potential energy = 12.5 Joules
falling distance = z sin 30 = .5 z
m g (.5 z) = 12.5
98 * .5 z = 12.5
z = 0.255 meters = 25.5 cm = total distance to stop
compresses spring .05 m
so hit spring at .255 - .05 = .25 meters slide distance
so fell .125 meters straight down before hitting spring
v = sqrt (2 g h) = sqrt (2 * 9.8 * .125) = 1.56 m/s
check my arithmetic

Answer 3

Oh, never mind the check. We agree :)