Question

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached to the other end. The spring is initially neither stretched nor compressed. A constant horizontal force of 51.0 N is then applied to the can, in the direction away from the post.
What is the speed of the can when the spring is stretched 0.400 m?
At the instant the spring is stretched 0.400 m, what is the magnitude of the acceleration of the block?

Answer 1

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2$\frac{m}{s^{2} }$

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

$K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=F$Δx=$(0.400m)$

W_{ext}=20.4 Nm

$U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2} \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}$

To determine, the acceleration we solve the following equation for a:

$F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }$