a 2.5 kg block is launched along the ground by a spring with a spring constant of 56 n/m. the spring is…
a 2.5 kg block is launched along the ground by a spring with a spring constant of 56 n/m. the spring is initially compressed 0.75 m. disregarding friction, how fast will the block move after the spring is released all the way and the block slides away from it? 3.5 m/s 4.1 m/s 13 m/s 16 m/s
Answer
Explanation:
Step1: Find elastic - potential energy
The formula for elastic - potential energy is $U = \frac{1}{2}kx^{2}$, where $k = 56$ N/m and $x = 0.75$ m. $U=\frac{1}{2}\times56\times(0.75)^{2}=\frac{1}{2}\times56\times0.5625 = 15.75$ J
Step2: Apply conservation of energy
Since there is no friction, the elastic - potential energy is converted entirely into kinetic energy $K=\frac{1}{2}mv^{2}$. We know $U = K$, $m = 2.5$ kg, and $U=15.75$ J. So, $\frac{1}{2}mv^{2}=15.75$ J. $v^{2}=\frac{2\times15.75}{2.5}=\frac{31.5}{2.5}=12.6$ $v=\sqrt{12.6}\approx 3.5$ m/s
Answer:
3.5 m/s