Physics Spring Work 3.docx

Physics Spring Work 3.docx - King Salmon Pulled by Eskimo...

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. King Salmon Pulled by Eskimo Example Given: f k = 0 m salmon & shed = 50.0 kg F = 1.20 × 10 2 N 1) What is the work done by the Eskimo on the sled if θ = 0.00 and the sled travels for 5.00 m? Step 1: Apply work → W = F × cosθ × ΔxStep 2: Plug in known values → W = (1.20 × 10 2 N) × cos(0.00 ) × (5.00 m) → W=(1.20×10 2 N)×(1)×(5.00m)→W=6.00×10 2 J
2) What is the work done by the Eskimo on the sled if θ = 30.0 and the sled travels for 5.00 m?Step 1: Apply work → W = F × cosθ × ΔxStep 2: Plug in known values → W = (1.20 × 10 2 N) × cos(30.0 ) × (5.00 m) → W = (1.20 × 10 2 N) × (0.87) × (5.00 m) → W = 5.20 × 10 2 J
 . 3) Let’s say f k ≠ 0 μ k (coefficient of kinetic friction) = 0.200 θ = 0 What is the work done on the sled by friction and the net work done on the sled by all external forces? Net-Work: Step 1: Apply 2 nd Law of Motion → F = m × aStep 2: Replace F with Ʃ Ʃ forces → F + f k + F g + n = m × aStep 3: Y-comp → F × sin(0 ) + f k × sin(180 ) + F g × sin(270 ) + n × sin(90 ) = 0 → F × (0) + f k × (0) + F g × (-1) + n × (+1) = 0 → -F g + n = 0 → n = F g Step 4: Replace F g with m × g → n = m × g → n = 50.0 kg × 9.80 m/s 2 n = 4.90 × 10 2 N
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Work done by Friction:Step 1: Apply work → W friction = F × cosθ × Δx Step 2: Change F → W friction = f k × cosθ × Δx → W friction = μ k × n × cosθ × ΔxStep 3: Plug in known values → W friction = 4.90 × 10 2 N × 0.200 × 5.00 N × cos(180 ) → W friction = 4.90 × 10 2 N × 0.200 × 5.00 N × -1 → W friction = -4.90 × 10 2 J
 Work Notes 3/15/17 What is the net work done on the block? Net work is the work done by the net force (F net ) W net = Fnet ×∆x ×cosθ = Fnet ×∆ x× cos ( 0 ) = Fnet ×∆ x Step 2: Use Newton’s Law of Motion = ma m | a || ∆ x | The net work is = | a || ∆ x | Monday March 20; Work and Spring Forces
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We assume that there is no friction down the slope only on the x EX: Ski V i = 0 Frictionless between A and B Kinetic friction coefficient B and C mu = 0.210 A) Find the skier’s speed at the bottom of the slope? W nc = KE + PE We have the force of gravity (conservative) and the Normal force (no work because the displacement is perpendicular to the surface cos 90 = 0; displacement and Force are in the same direction. We have conservation of mechanical energy) 1 2 mv a 2 + m g y a = 1 2 m v b 2 + mg y b 0 + ( 9.8 m s 2 ) ( 20.0 m ) = 1 2 v b 2 + 0 2 ( 9.8 m s 2 ) ( 20.0 m ) ¿ ¿ v b = ¿
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B) How far does the skier travel on the horizontal surface before coming to rest? What is x v b = 19.8 m / s V c = 0 mu = 0.210 W nc = ∆ KE + ∆PE The forces that are acting on the skier are frictional force, force of gravity (conservative), and normal force (does no work). Since the skier is moving to the right only the nonconservative forces acting on the skier is frictional force. W nc = W fk = | f k | | d | cosθ ¿ | f k | | d | cos ( 180 ° ) mu k | n | d Apply Newton’s 2nd Law of motion | n | = | F g | =− mu k | Fg | d =− mu k mg d ¿ ∆ KE + ∆ PE ¿ 1 2 m v c 2 1 2 m v b 2 + mg y c mg y b 1 2 m v b 2 ¿ mu k mgd = 1 2 mv b 2 ¿ d = ( v b 2 ) 2 mu k ( g ) ( 19.8 m s ) 2 2 ( 0.21 ) ( 9.80 ) = 95.2 m Spring Force Potential energy stored in a Spring Involves the spring constant, k (characteristic of a given spring; not always the same)
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