A 5.0 kg block is placed near the topof a frictionless ramp, which makes an angl

Question

A 5.0 kg block is placed near the topof a frictionless ramp, which makes an angle of 24° to the horizontal. A distance d = 1.3m away from the block is an unstretched spring with k =3300 N/m. The block slides down the rampand compresses the spring. Find the maximum compression of thespring (X max).
I made an attempt by making the potential energy of the blockequal to the block's kinetic energy and solved for X: mass * gravity * change in height = .5 * k *X^2 5.0 kg * 9.8m/s^2 * 1.3 * sin (24) = 0.5 * 3300 * X^2
X = 0.1253 m But this values of X is wrong. Can someone please explain howto do this problem properly?

Explanation / Answer

  From law of conservastion of energy    mgsin (d+x) = 1/2kx2      gsin d + gsin x = 0.5x2      0.5(3300) x2 = (9.8 m/s2) (sin240) (1.3)+  (9.8 m/s2) (sin240) x     1650   x2 =5.2 + 3.986x        1650   x2  - 3.986x - 5.2 =0         x =0.057 m or5.7cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---