Physics Problem: Work on a Rescue Sled on a Slope
Question:
Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 100.1 kg, down a 60.0° slope at constant speed. The coefficient of kinetic friction between the sled and the snow is 0.100.
Determine:
- (a) Work done by friction as the sled moves 30.4 m along the hill.
- (b) Work done by the rope on the sled over this distance.
- (c) Work done by gravity on the sled.
- (d) Net work done on the sled.
Step-by-Step Solution:
Step 1: Work Done by Friction (Wf)
N = mg × cos(θ)
fk = μk × N
Wf = −fk × d
Given:
- m = 100.1 kg
- g = 9.8 m/s²
- θ = 60.0°
- μk = 0.100
- d = 30.4 m
Calculation:
N = 100.1 × 9.8 × cos(60°) = 980.98 × 0.5 = 490.49 N
fk = 0.100 × 490.49 = 49.05 N
Wf = −49.05 × 30.4 = −1490.12 J
Step 2: Work Done by the Rope (Wt)
Wt = T × d × cos(0°) = T × d
Assuming: T = 400 N
Wt = 400 × 30.4 = 12160 J
Step 3: Work Done by Gravity (Wg)
Wg = −mg × d × cos(120°)
cos(120°) = −0.5
Calculation:
Wg = −100.1 × 9.8 × 30.4 × 0.5 = −14900.88 J
Step 4: Net Work Done (Wnet)
Wnet = Wt + Wf + Wg
Wnet = 12160 + (−1490.12) + (−14900.88) = −1230.99 J
Final Answers:
- (a) Work by friction: −1490.12 J
- (b) Work by rope: 12160 J
- (c) Work by gravity: −14900.88 J
- (d) Net work done: −1230.99 J