A MAN GRASP A ROPE 6M ABOVE THE GROUND, FROM HIS HAND THE ROPE GOES STRAIGHT UP TO A PULLEY 36M ABOVE THE GROUND AND THEN STRAIGHT DOWN TO THE WEIGHT RESTING ON THE GROUND. IF THE MAN, HOLDING FAST TO A ROPE AND KEEPING HIS HAND 6M ABOVE THE GROUND WALKS AWAY AT A CONSTANT RATE OF 10M/S, HOW FAST IS THE WEIGHT RISING:

Question

A MAN GRASP A ROPE 6M ABOVE THE GROUND, FROM HIS HAND THE ROPE GOES STRAIGHT UP TO A PULLEY 36M ABOVE THE GROUND AND THEN STRAIGHT DOWN TO THE WEIGHT RESTING ON THE GROUND. IF THE MAN, HOLDING FAST TO A ROPE AND KEEPING HIS HAND 6M ABOVE THE GROUND WALKS AWAY AT A CONSTANT RATE OF 10M/S, HOW FAST IS THE WEIGHT RISING:

I. 1 SECOND AFTER THE MAN STARTS TO WALK
II. WHEN THE WEIGHT IS 20M ABOVE THE GROUND.

Answer 1

To determine how fast the weight is rising, we can use the concept of related rates. We need to find the rate at which the distance between the man's hand and the weight is changing.

First, let's consider the situation at 1 second after the man starts to walk. At this point, the man has walked a distance of 10m (10m/s x 1s) away from the weight. Since the rope is straight, the length of the rope from the man's hand to the weight remains constant at 36m.

To visualize the situation, let's draw a right triangle with the horizontal distance between the weight and the man as the base (let's call it x), the height of the triangle as the distance between the man's hand and the ground (6m), and the hypotenuse as the length of the rope (36m).

Using the Pythagorean theorem, we can determine the value of x:

x^2 + 6^2 = 36^2
x^2 + 36 = 1296
x^2 = 1296 - 36
x^2 = 1260
x = √1260
x ≈ 35.49m

Now, let's find the rate at which x is changing. We can do this by differentiating the equation with respect to time:

d(x^2)/dt = d(1260)/dt
2x(dx/dt) = 0
dx/dt = 0 / (2x)
dx/dt = 0

So, 1 second after the man starts to walk, the weight is not rising.

Next, let's consider the situation when the weight is 20m above the ground. At this point, the length of the rope from the pulley to the weight is 56m (36m + 20m).

Again, using the Pythagorean theorem, we can find x:

x^2 + 6^2 = 56^2
x^2 + 36 = 3136
x^2 = 3136 - 36
x^2 = 3100
x = √3100
x ≈ 55.64m

Differentiating the equation with respect to time, we have:

2x(dx/dt) = 0
dx/dt = 0 / (2x)
dx/dt = 0

So, when the weight is 20m above the ground, it is not rising either.

Therefore, in both cases, the weight is not rising.