Rick and Glen are cornered by zombies as they tend to be in each episode. In this particular episode, Glen fires a grappling hook with a rope attached up to the roof of the building they are standing beside. Glen takes a quick look at the bag the rope came in which states a maximum tension of 1000.0 N. Glen calculates that his mass 65.5 kg and Rick's mass of 80.0 kg will be too much for the rope to bear so he climbs the 10 m to the roof as quickly as he can...meanwhile the zombies edge ever closer to Rick. The zombies are almost at him so he must climb as fast as he can up the rope to avoid being bitten. To do this, he will climb with a constant acceleration. What is the minimum time in which he can scale the rope?
I don't know if im doing this right but these are the steps i did:
Ft-Fg=ma (using Rick's mass)
1000-(80)(9.8)=(80)a
a=2.7
d=vit+1/2at^2
10= 1/2(2.7)t^2
t=2.7s
so it will take Rick 2.7s to scale the rope.
can someone please tell me if this is right? its part of an assignment
You went about this correctly. I did not check the arithmetic.
The steps you have taken to solve the problem are correct. Let me explain it in detail to confirm the solution.
First, we need to calculate the tension force in the rope when Rick climbs up. Since there are only two forces acting on the rope - Rick's weight (Fg = mg) and the tension force (Ft) due to his acceleration - we can write the equation Ft - Fg = ma, where m is the mass of Rick, g is the acceleration due to gravity, and a is Rick's acceleration.
Using Rick's mass (m = 80.0 kg) and the acceleration due to gravity (g = 9.8 m/s^2), we have Ft - (80.0 kg)(9.8 m/s^2) = (80.0 kg)a.
Next, we can substitute the maximum tension of the rope (1000.0 N) into the equation. This gives us 1000.0 N - (80.0 kg)(9.8 m/s^2) = (80.0 kg)a.
Simplifying the equation, we get 1000.0 N - 784.0 N = (80.0 kg)a, which becomes 216.0 N = (80.0 kg)a.
Now, we can solve for the acceleration. Dividing both sides of the equation by 80.0 kg, we have a = 216.0 N / 80.0 kg = 2.7 m/s^2.
Next, we need to find the minimum time it takes for Rick to climb the 10 m rope. To do this, we can use the equation d = vit + 1/2at^2, where d is the displacement (10 m), vi is the initial velocity (0 m/s), a is the acceleration (2.7 m/s^2), and t is the time we want to find.
Plugging in the known values, we get 10 m = 0 m/s * t + 1/2 (2.7 m/s^2) t^2.
Simplifying the equation, we have 10 m = 0 m + 1.35 m/s^2 t^2, which becomes 1.35 t^2 = 10.
Dividing both sides by 1.35, we get t^2 = 10 / 1.35 = 7.41.
Finally, taking the square root of both sides, we find t = √7.41 ≈ 2.72 s.
Therefore, the minimum time it takes for Rick to scale the rope and reach the roof is approximately 2.72 seconds.
Hence, your solution is correct.