How Fast Will the Climber Accelerate and How Long Until the Rock Falls?

[orange03]

A 75 kg climber finds himself dangling over the edge of an ice cliff, as shown in the figure below. Fortunately, he's roped to a 980 kg rock located 51 m from the edge of the cliff. Assume that the coefficient of kinetic friction between rock and ice is 5.5×10−2. What is his acceleration, and how much time does he have before the rock goes over the edge? Neglect the rope's mass. What is his acceleration? How much time does he have before the rock goes over the edge?

+Ft-Fr=max
+Ft-mg=may

ax=(mg-uk(Mg))/(M-m)
I keep getting ax= 0.228 m/s^2 but that's the answer and I do not know what I am doing wrong. Help please!

 

[tiny-tim]

Welcome to PF!

Hi orange03! Welcome to PF!

ax=(mg-uk(Mg))/(M-m)

Why (M-m)?

 

[tomcue]

I would first make sure that all units are consistent and that all necessary variables are included in the calculation. From the given information, we know that the climber has a mass of 75 kg and is connected by a rope to a rock with a mass of 980 kg. The distance from the edge of the cliff to the rock is 51 m, and the coefficient of kinetic friction between the rock and ice is 5.5×10−2. The only variable that we do not have is the acceleration, which we will solve for.

Using the equations for Newton's Second Law (ΣF=ma) and the force of friction (Ff=μN), we can set up the following equations:

ΣFx = Fapp - Ff = ma
ΣFy = N - mg = 0

Since the climber is not moving horizontally, we can set ΣFx = 0. This means that Fapp = Ff, or in other words, the force applied by the climber must be equal to the force of friction between the rock and ice. We can rewrite this as:

Fapp = μN = μ(mg) = (5.5×10−2)(75 kg)(9.8 m/s^2) = 40.95 N

Now, we can plug this value for Fapp into our original equation for ΣFx and solve for the acceleration, a:

0 = Fapp - Ff = ma
0 = 40.95 N - μN = (75 kg)a
a = (40.95 N)/(75 kg) = 0.546 m/s^2

So, the climber's acceleration is 0.546 m/s^2.

To find the time before the rock goes over the edge, we can use the equation y = y0 + v0t + 1/2at^2, where y0 is the initial height (51 m) and v0 is the initial velocity (0 m/s). We can solve for t when y = 0, which represents the point where the rock reaches the edge of the cliff.

0 = 51 m + 0 m/s + 1/2(0.546 m/s^2)t^2
t = √(2(51 m)/(0.546 m/s^2)) = 10.39 seconds

Therefore,