a 1210 kg roller coaster car is moving 6.33 m/s. as it approaches the station, brakes slow it down to 2.38 m/s over a distance o

Question

3 years ago

12

f 4.20 m. what is the magnitude of the force applied by the brakes?(unit=N) PLEASE HELP

2 answers:

Ratling [72] · Answer 1 · 2022-04-25T02:08:30+03:00

8 0

Answer:

4,955.9 N

Explanation:

1/2 (m*v^2) = 1/2 (1210 x 6.33^2) = 24,241.7 J.

1/2 (1210 x 2.38^2) = 3,427 J.

(24,241.7 - 3,427) = 20,814.7J. of work done by the brakes.

(20,814.7/ 4.2m) = brake force of 4,955.9 N.

Stels [109] · Answer 2 · 2022-04-25T01:06:45+03:00

Answer:

4960 N

Explanation:

First, find the acceleration.

Given:

v₀ = 6.33 m/s

v = 2.38 m/s

Δx = 4.20 m

Find: a

v² = v₀² + 2aΔx

(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)

a = -4.10 m/s²

Next, find the force.

F = ma

F = (1210 kg) (-4.10 m/s²)

F = -4960 N

The magnitude of the force is 4960 N.