The 50-lb block rests on the rough surface for which the coefficient of kinetic friction is 14k = 0.2. A force F = (40+ s2) lb, where s is in ft, acts on the block in the direction shown. The spring is originally unstretched (s = 0) and the block is at rest. (Figure 1) ▾ Part A Determine the power developed by the force the instant the block has moved s = 1.5 ft. Express your answer in units of horsepower to three significant figures. Figure 30° k = 20 lb/ft www <1 of 1 > 1 ΑΣΦ 11 vec W = Submit Request Answer Provide Feedback ? hp

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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The 50-lb block rests on the rough surface for which the coefficient of kinetic friction is
μk = 0.2. A force F = (40+ s²) lb, where s is in ft, acts on the block in the direction
shown. The spring is originally unstretched (s = 0) and the block is at rest. (Figure 1)
Part A
Determine the power developed by the force the instant the block has moved s = 1.5 ft.
Express your answer in units of horsepower to three significant figures.
Figure
F.
k = 20 lb/ft
30°
1 of 1
W =
VAΣ T
ΜΕ ΑΣΦ
vec
Submit
Request Answer
Provide Feedback
?
hp
Transcribed Image Text:The 50-lb block rests on the rough surface for which the coefficient of kinetic friction is μk = 0.2. A force F = (40+ s²) lb, where s is in ft, acts on the block in the direction shown. The spring is originally unstretched (s = 0) and the block is at rest. (Figure 1) Part A Determine the power developed by the force the instant the block has moved s = 1.5 ft. Express your answer in units of horsepower to three significant figures. Figure F. k = 20 lb/ft 30° 1 of 1 W = VAΣ T ΜΕ ΑΣΦ vec Submit Request Answer Provide Feedback ? hp
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