questions about force?

Question

1) A race car has a mass of 718 kg. It starts from rest and travels 44.0 m in 3.0 s. The car is uniformly accelerated during the entire time. What net force is exerted on it?
N=?

2)Your mass is 80 kg. You stand on a bathroom scale in an elevator on Earth.

(a) What force would the scale exert when the elevator moves up at a constant speed?
N=?
(b) What force would the scale exert when it slows at 2.2 m/s2 while moving upward?
N=?
(c) What force would the scale exert when it speeds up at 2.2 m/s2 wile moving downward?
N=?
(d) What force would the scale exert when it moves downward at a constant speed?
N=?
(e) What force would the scale exert when it slows to a stop at a constant magnitude of acceleration?



3) A boy exerts a 38 N horizontal force as he pulls a 54 N sled across a cement sidewalk at constant speed. What is the coefficient of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance.

Answer 1

The three equations you need for dealing with forces are

force F = mA
speed V = At + Vi (Vi = initial speed)0
distance D = 1/2At^2 + ViT

With these three you can solve anything.

1) Lay out the equations

F = mA; you know m, you want F so you need A
V = At (Vi = 0); A = V/t; you know t so you need V
D = 1/2 At^2; replace A by V/t; D = 1/2 (V/t) t^2
So D = 1/2 V^t; V = 2D/t = 88m/3s = 29.3m/s

A = V/t = 29.3m/s / 3s = 9.8 m/s^2
F = mA = 718 kg x 9.8m/s^2 = 7036 N

2) Assume g = earth's acceleration due to gravity = 10m/s^2

a) N = 80 kg x 10m/s^2 = 800 N
b) N = 80 kg x (10 + 2.2)m/s^2 = 976 N
c) N = 80 kg x (10-2.2)m/s^2 = 624 N
d) N = 80 kg x 10m/s^2 = 800 N
e) more than 800N and without knowing the magnitude, it is impossible to compute. Formula would be

N = 80 kg (10 + a)m/s^2

3) u = 38/54 = 0.70

Answer 2

I thought I answered that question already .
Okay
1. Let's use famous F=ma

Average a = V/t
so
F=m v/t= 718 x 44.0/3.0=
F=10530 N

2.
(a ) did you say constant speed? then W=mg

(b) 2.2 m/s2 mowing upward then
F=W+Fe= mf +ma=m(g+a)

(c)2.2 m/s2 mowing downward then
F=W-Fe= mf +ma=m(g-a)

(d) same as (a)

(e) constant negative acceleration then it is the ame as (c)


3. Force of pull equal to force of friction
F=f
F=uN
u=F/N=38/54=0.704

Have fun again!