Chapter 8: Q128P (page 212)

A70 kgfirefighter slides, from rest,4.3 mdown a vertical pole. (a) If the firefighter holds onto the pole lightly, so that the frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor? (b) If the firefighter grasps the pole more firmly as she slides so that the average frictional force of the pole on her is500 Nupward, what is her speed just before reaching the ground floor?

Short Answer

Expert verified

(a) The speed of the firefighter on the ground floor is $v = 9.2 m / s$ .

(b) The speed of the firefighter at the ground floor when frictional force 500 N is acting upward is $v_{f} = 4.8 m / s$ .

Step by step solution

Given data:

The mass of the firefighter, m = 70 kg

The height of the pole, $h_{0} = 4.3 m$

The initial velocity, $v_{0} = 0 m / s$

The frictional force. f = 500 N

To understand the concept:

Use the concept of conservation of mechanical energy to find the speed of the firefighter at the ground floor. In the second case, use work done by frictional force related to change in mechanical energy and solve for speed at the ground floor.

Formulae:

The mechanical energy is,

$M E = K E + P E = \frac{1}{2} m v^{2} + m g h$

The work done is,

$W_{f} = M E_{f} + M E_{i}$

(a) The speed of the firefighter at the ground floor:

Using the conservation of mechanical energy equation, you can find the speed of a firefighter at the ground floor.

$\frac{1}{2} m v^{2} + m g h = \frac{1}{2} m v_{i}^{2} + m g h_{i}$

Where, h is the final height, which is zero at ground floor and g is the acceleration due to gravity having a value $9.8 m / s^{2}$ . Initial velocity $(v_{i})$ is zero as she slides from rest.

Mass is common at both sides; you can cancel it out.

$\frac{1}{2} v^{2} + g h = \frac{1}{2} v_{i}^{2} + g h_{i}$

Plugging the values, you get

$\frac{1}{2} v^{2} + (9.8 m / s^{2} \times 0) = \frac{1}{2} {(0)}^{2} + (9.8 m / s^{2} \times 4.3 m) \frac{1}{2} v^{2} = (4214 m^{2} / s^{2}) v^{2} = 84.28 m^{2} / s^{2} v = 9.2 m / s$

Hence, the speed of the firefighter on the ground floor is $v = 9.2 m / s$ .

(b) The speed just before reaching the ground floor:

Speed of the firefighter at the ground floor when frictional force 500 N is acting upward.

When there is frictional force opposing the motion, work done by this frictional force will decrease the speed of the firefighter. You can write,

$W_{f} = (\frac{1}{2} m v_{f}^{2} + m g h) - (\frac{1}{2} m v_{i}^{2} + m g h_{i}) - f \times h_{i} = (\frac{1}{2} m v_{f}^{2} + m g h) - (\frac{1}{2} m v_{i}^{2} + m g h_{i})$

Negative sign is for force, and displacement is opposite. Plugging the values, you get,

$- 500 N \times 4.3 m = (\frac{1}{2} (70 k g) v_{f}^{2} + (70 k g) (9.8 m / s^{2}) (0)) - (\frac{1}{2} (70 k g) {(0)}^{2} + (70 k g) (9.8 m / s^{2}) (4.3 m)) - 2150 N . m = ((35 k g) v_{f}^{2} + 0) - (0 + 2, 949.8 N . m) (35 k g) v_{f}^{2} = (2, 949.8 = 2150) N . m (35 k g) v_{f}^{2} = 799.8 N . m v_{f}^{2} = 22.8514 m^{2} / s^{2} v_{f} = 4.8 m / s$