Physics Spring Work 3.docx

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

• 19

This preview shows page 1 - 5 out of 19 pages.

. 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

Subscribe to view the full document.

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
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 = ¿

Subscribe to view the full document.

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)
You've reached the end of this preview.
• Spring '08
• Cavatto

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern