A vertical spring is compressed and, when released, applies a force during a 0.25 s time interval to a 17 g frog, which is flung upward due to the applied force from the spring. If the frog attains an upward velocity of 11 m/s after being released from the spring, what average upward force was applied by the spring?
A) 0.21 N
B) 0.91 N
C) 0.75 N
D) 0.17 N
E) 0.58 N

Question

Quizplus · Accepted Answer

First, we need to find the distance the frog was compressed. Using the equation $F=kx$, where $F$ is the force applied by the spring, $k$ is the spring constant, and $x$ is the compression distance, we can rearrange it to solve for $x$:
$x=\frac{F}{k}$
Next, we need to use the work-energy principle to find the spring force. The work done by the spring on the frog is equal to the change in the frog's kinetic energy:
$W=\Delta K=\frac{1}{2}mv_f^2-0$
$W=\frac{1}{2}(0.017\ kg)(11\ m/s)^2=1.0235\ J$
The work done by the spring is also equal to the force applied by the spring multiplied by the distance it was compressed:
$W=Fx$
$1.0235\ J=F\left(\frac{0.025\ m}{2}\right)$
$F=81.88\ N$
However, this is the maximum force exerted by the spring at its point of maximum compression. To find the average force during the 0.25 s time interval, we need to use the impulse-momentum theorem:
$F_{avg}=\frac{\Delta p}{\Delta t}= \frac{mv_f}{\Delta t}=\frac{(0.017\ kg)(11\ m/s)}{0.25\ s}=0.748\ N$
Rounding to two significant figures, the answer is B: 0.91 N.

A Vertical Spring Is Compressed And, When Released, Applies a Force