Notes to help with Unit VII Worksheet 3b

Formulas:

E_g = mgh work = area under curve of F vs Dx graph = FDx (for a constant force)

E_k = ½ mv² work = area under curve of F vs Dx graph = ½ FDx (for a variable force)

E_el = ½ kDx²

Sample problem #1: Find the work done against friction.

E_i = E_f + work out so work done by friction = E_i – E_f = 9800 j – 6050 j = 3750 j.

If you define the system as the cart, earth and track, you could think of the problem this way:

No work in or out so E_i = E_f and 9800 j = 6050 j + E_diss

E_diss = work against friction = 9800 j - 6050 j = 3750 j

Sample problem #2: A 10.0 kg pumpkin initially at rest is lifted 5.0 m with a force of 200.0 N, straight up. What is its speed when you let it go?

Define the system: pumpkin, earth

E_i + work in = E_f and initial energy = 0 so work in = E_f

Work done on pumpkin by person: work = FDx = (200.0 N)(5.0 m) = 1000 j

Final energy: E_g = mgh = (10.0 kg)(5.0 m)(9.8 m/s²) = 490 j

E_k = 1000 j - 490 j = 510 j

Finding velocity: E_k = ½ mv²

510 j = ½ (10.0 kg) v²

v = sqrt [(2)( 510 j) / (10.0 kg)] = 10.1 m/s

Sample problem #3: A 0.50 kg air cart moving on a (relatively) frictionless air track at 1.0 m/s hits a spring with a spring constant of 22.0 N/m. By how much is the spring compressed when the air cart come to a stop?

initial energy: E_k = ½ mv² = ½ (0.50 kg)( 1.0 m/s) ² = 0.25 j

no work in or out so No work in or out so E_i = E_f and E_g = E_el = 0.25 j

E_el = ½ kDx² so Dx = sqrt (2 E_el / k) = sqrt [(2)(0.25 j ) / 22 N/m)] = 0.15 m