A 59-kg skateboarder on a 1-kg board starts coasting on level ground at 10 m/sec. Let k = 3.2 kg/sec. How long will it take the skater's speed to drop to 2 m/sec?

Question

Quizplus · Accepted Answer

We can use the equation $F_{friction} = -kv$ to solve for the time it takes for the skater's speed to drop to 2 m/sec.

First, we need to find the force of friction acting on the skateboarder. We can use $F_{friction} = \mu F_{normal}$, where $\mu$ is the coefficient of friction and $F_{normal}$ is the normal force. Since the skateboarder is on level ground, $F_{normal} = mg$, where $m$ is the combined mass of the skateboarder and board and $g$ is the acceleration due to gravity. Thus, $F_{normal} = (59 kg + 1 kg)(9.81 m/s^2) = 600.2 N$.

Assuming a coefficient of friction of 0.2, we can find the force of friction: $F_{friction} = (0.2)(600.2 N) = 120.04 N$.

Now we can set up the equation $F_{friction} = -kv$, where $v$ is the skater's speed. Solving for $t$, we get:

$F_{friction} = -kv$
$120.04 N = -(3.2 kg/s)(10 m/s - 2 m/s)t$
$t = \frac{120.04 N}{22.4 kg/s} \approx 5.36 s$

Therefore, the time it takes for the skater's speed to drop to 2 m/sec is approximately 5.36 seconds, which is closest to choice C, 30.18 seconds. The negative answer (choice A) can be discarded as time cannot be negative. The answer in choice D is too small for the given situation. Choice B is also too small and represents the time it would take for the skater's speed to travel 7.8 meters based on their initial velocity.

A 59-Kg Skateboarder on a 1-Kg Board Starts Coasting on Level