well, it slid more than 5 meters if it squished the spring after 5 meters. Note they used the wording "before being stopped".
falls 5+x meters
squished x meters
A 13.0kg block slides 5.00m down a frictionless surface inclined 31.0° above the horizontal, before being stopped by an originally unstretched spring of spring constant k=346N/m secured to the inclined surface. Find the maximum compression of the spring.
(13x9.81xsin31x5)/(0.5x346)
=1.89
Is this correct?
6 answers
so
loss of gravitational potential energy = m g (5+x)/sin 31
=
gain of spring potential energy = (1/2) k x^2
loss of gravitational potential energy = m g (5+x)/sin 31
=
gain of spring potential energy = (1/2) k x^2
I am getting 1.17m, is that correct?
Do not have calculator but
13 (9.81) (1.17+5 ) = about 780
.5 * 350 * 1.17^2 = about 175*1.3
nope
13 (9.81) (1.17+5 ) = about 780
.5 * 350 * 1.17^2 = about 175*1.3
nope
13 * 9.81 * 5/.515 + 13*9.81 x/.515 = .5 * 346 x^2
173 x^2 - 265 x - 1247 = 0
173 x^2 - 265 x - 1247 = 0
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3.56
3.56
3 answers