A 0.454-kg block starts from rest and slides 3.05 m down a frictionless plane inclined at 53º to the horizontal.At the bottom it slides 9.14 m over a horizontal frictionless plane before compressing a spring (k = 14.7 N/m)a distance x and coming momentarily to rest.The value of x is approximately
A)0.215 m
B)1.52 m
C)2.44 m
D)1.83 m
E)1.21 m

Question

Quizplus · Accepted Answer

First, we can use conservation of energy to find the speed of the block at the bottom of the incline:

$PE_i = KE_f$

$mgh = \frac{1}{2}mv_f^2$

$3.05\sin(53^\circ)(9.8)m = \frac{1}{2}(0.454)v_f^2$

$v_f \approx 3.50$ m/s

Next, we can use conservation of energy again to find how much the spring is compressed:

$KE_i + PE_i = KE_f + PE_f + \frac{1}{2}kx^2$

$\frac{1}{2}mv_f^2 = \frac{1}{2}kx^2$

$x \approx 1.83$ m (rounded to two significant figures)

A 0454-Kg Block Starts from Rest and Slides 3