Dynamics - 2 - 2nd Law Single-Body | High School

University:
High School
Course:
High School Physics
Academic year:

2024
Views:

414

Pages:

10
Author:

1 1

4.2 Newton’s nd 2 Law (cont.) Single-body problems Ex. #1 A force of 10.0N is applied at an angle of 30.0o with the horizontal to a 2.00kg block resting on a surface. The force of friction is 0.500N. F=10N 30o Ffric=0.5N 2.0kg What is the acceleration of the block? Fnet = Fapplied − Ffriction Fnet = ma F is diagonal - must find horizontal component (effective force) 10N 30o ? cos 30 = 10 Fapp = 8.66N Fnet = 8.66 − 0.500N Fnet = 8.16N Fnet = ma 8.16 = ( 2.0)a a = 4.08m / s 2 Ex. #2 A student applies various weights to a spring & obtains the following results: 20.0 Force (N) 10.0 0.25 0.50 Extension (m) The student attaches it to a 5.00kg cart on a rough flat surface. She pulls on the spring and it extends 62.5cm while accelerating 3.20m/s2. 3.20m/s2 5.0kg 0.625m Fapp What is the magnitude of the frictional force between the cart & the rough surface? ma = 25.0 − Ffriction Fnet = Fapplied − Ffriction(5.00)(3.20) = 25.0 − F spring F = kx recall, k = slope 20.0 − 10.0 k= 0.50 − 0.25 k friction F friction = 9.0N = 40.0N / m Fspring = (40.0)(0.625) Fapp = 25.0N Ex. #3 A car with a mass of 1000.0 kg, moving at a speed of 30.0 m/s brakes to rest over a distance of 100.0 metres. What is the magnitude of the force of friction which causes the car to stop? m = 1000.0kg v1 = 30.0m/s v2 = 0m/s d = 100.0m Ffric = ? 2 (v2 ) = (v1 ) + 2ad 2 (0) = (30) + 2(a)(100) 2 2 a = −4.50m / s Fnet = ma 2 Ffric = 4.50 •10 N 3 Fnet = (1000)( −4.50) Fnet = −4500N Fnet = Fapp − Ffric But Fapp = 0, so Fnet = - Ffric Ex. #4 A 1.3 g paintball takes 8.7  10-3 s to exit the barrel of a paintball gun which is 0.35 m long. The acceleration is constant. What force does the explosion of the CO2 exert on the paintball? (neglect friction) m = 1.3g d = 0.35m t = 8.7 x 10-3 s v1 = 0m/s Fapp = ? 1 2 d = v1t + at 2 1 0.35 = (0)(8.7 • 10 ) + (a )(8.7 • 10−3 ) 2 2 −3 a = 9248.249m / s Fnet = ma 2 But Ffric = 0, so Fnet = Fapp Fapp = (0.0013)(9248.249) Fapp = 12N HW: pg 314 #3, 5, 7, 9, 11 Quantum