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?\n3.5 m/s\n4.1 m/s\n13 m/s\n16 m/s
Answer
Explanation:
Step1: Calculate 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 into kinetic energy $K=\frac{1}{2}mv^{2}$. We know $U = K$, $m = 2.5$ kg, and $U = 15.75$ J. $\frac{1}{2}mv^{2}=15.75$ J, so $v^{2}=\frac{2\times15.75}{m}$. Substituting $m = 2.5$ kg, we get $v^{2}=\frac{2\times15.75}{2.5}=\frac{31.5}{2.5}=12.6$. Then $v=\sqrt{12.6}\approx 3.5$ m/s
Answer:
3.5 m/s