A 500-kg car is pulled with a force of 120 N, but it accelerates at just 0.1 m/s^2. What force of friction must be acting on the car?
回答 (3)
2018-08-10 4:46 pm
Let f (N) be the frictional force.
F = m a
120 - f = 500 × 0.1
120 - f = 50
Frictional force, f = 70 N
F = m a
120 - f = 500 × 0.1
120 - f = 50
Frictional force, f = 70 N
2018-08-10 11:25 pm
find net force
Fnet - m a = 500 kg (0.1 m/s/s) = 50 N
find friction
Fnet = Fa - Ff
50 N = 120 N - Ff
Ff = 120 N - 50 N = 70 N
When you get a good response,
please consider giving a best answer.
This is the only reward we get.
Fnet - m a = 500 kg (0.1 m/s/s) = 50 N
find friction
Fnet = Fa - Ff
50 N = 120 N - Ff
Ff = 120 N - 50 N = 70 N
When you get a good response,
please consider giving a best answer.
This is the only reward we get.
2018-08-11 8:12 am
The net force on the car is the applied force minus the friction force.
120 – Ff = 500 * 0.1
Ff = 120 – 50 = 70 N
120 – Ff = 500 * 0.1
Ff = 120 – 50 = 70 N
收錄日期: 2021-04-24 01:06:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180810083532AAaPxVr