Force and Motion, horizontal forces

[matadorqk]

Homework Statement

20. A 873-kg (1930 lb) dragster, starting from rest, attains a speed of 26.3 m/s (58.9mph) in 0.59 s.
-----a. Find the average acceleration of the dragster during this time interval.
My solution:
Given variables:
M=873kg
Vo=0
Starting time: 0
V=26.3m/s
T=0.59
Formula to be used: V=Vo + at converted into A=v-vo/t
It equals:
a= 26.3/0.59=44.6m/s^2
-----b. What is the magnitude of the average net force on the dragster during this time?
My solution:
Given variables:
m=873 kg
a=44.6m/s^2
F net= ma
F net = 38,935.8 N
F net = 39,900N
-----c. Assume that the driver has a mass of 68kg. What horizontal force does the seat exhert on the driver?
Here is where i have a problem..[B/]
I would assume I have to do:
F net=ma using 68kg as my M and 44.6m/s as my A to solve for the force exherted by the seat, which would equal 1,788.4N (Horizontal Force seat exherts) Yet I don't know if that's done correctly.. I would love someone to revise over the work, especially C to see if I did it correctly..

 

[robb_]

Looks good to me.

 

[matadorqk]

Awesome,

Well that's great, then I guess that means I am doing the rest of the homework right :P! Hehe thanks

 

[matadorqk]

Help again i think

I think I am stuck again,

37. a sled of masss 50.0kg is pulled along flat, snow-covered ground. The static friction coefficient is 0.30, and the kinetic friction coefficient is 0.10
-----a. What does the sled weigh?
Im assuming they mean force, so it would be 50kg x 9.8 to convert to N,so 490N, and get its "weight"? I think this is how it is, although it seems too easy, if anyone would revise this id be grateful :)
-----b. What force will be needed to start the sled moving?
My given values are:
m=50kg
us(coefficient of static friction)=0.3
uk(coefficient of kinetic friction)=0.1
Fp=uk*m*g which is Fp=0.1*50*9.8= 49N
So is the force needed to start 49 Newtons?

 

[robb_]

looks good so far

 

[matadorqk]

I think the "So is the force needed to start 49 Newtons?" is wrong, because now that I think of it wouldn't I insttead use the Fs= us x mg which would equal 147 N (which is 0.3 x 50 x 9.8)..

 

[robb_]

You need to overcome static friction to get the object to move.

 

[matadorqk]

oh so then its like.. above 147? like 148+? Cause static friction is 147N, so yeah, I guess 148N would be overcoming it

thanks again for the fast reply