# Homework Help: Force and Acceleration/Force of Friction and Air Resistance. 5 Basic Questions.

1. Oct 25, 2009

### Underdog_85

Q1: Assuming friction force to be negligibly small, what will be the acceleration in m/s^2 of a 45kg skater when propelled by a force of 250N?
My attempt
Data:

F-250N
m-45kg
a-m/s^2

Formula:
F=ma
a=F/m
a=250N/45kg
a=5.555
a=5.6m/s^2
Q2: a 0.5kg mass on a horizontal frictional free table is accelerated by a force from a string passing over a pulley and attached to a mass of 1.5kg hanging vertically.

My attempt
Data:

m-0.5kg
hanging m-1.5kg

formula:
F=F1+F2
F=0.5kg+1.5kg=2.0kg
F=ma
F=2.0kg(9.80m/s^2)
F=19.6 rounded to 20N

My attempt at B)
Data:

F=19.6N
m=2kg
a=?m/s^2?

Formula:
F=ma
a=F/m
a=19.6N/2kg
a=9.8 rounded to 10m/s^2
Q3: A 1200kg truck is moving at 108km/h and stops within a distance of 91m
a)What is the acceleration?

My attempt
Data:

m-1200kg
Vi-108km/h--->converted to 30m/s
Vf-0
s-91m

Formula:
2as=Vf^2+Vi^2
a=Vi^2/2s
a=(30m/s)^2/2(91m)
a=900m^2/s^2/182m
a=-4.9m/s^2(negative because it decelerated)

My attempt at part B)What is the coefficient of friction?
Formula:

Ff=uFn
u=Ff/Fn
u=4.9m/s^2/1200kg
u=0.00408
Q4: you are pulling a 3.0slug crate and a sliding friction is known to be 55lb. what is the acceleration o the crate when the pulling force is?
a)55lb, b)75lb and c)105lb

My attempt at a)
Date:

m: 3.0Slug
F:55lb
a:?m/s^2?

Formula:
F=ma
a=F/m
a=55lb/3.0slug
a=18ft/s^2

My attempt at part b)
Data:

m: 3.0slug
F: 75lb
a: ??

formula:
F=ma
a=F/m
a=75lb/3.0slug
a=25ft/s^2

My attempt at part c)
Data:

m: 3.0slug
F:105lb
a-??

Formula:
F=ma
a=F/m
a=105lb/3.0slug
a=35ft/s^2
Q5: During a training exercise, Brian who weighs 75kg skydives and parachutes from a stationary helicopter. What is the acceleration when the air drag is a)500N, b)1500N, and c)Equal to his weight

My attempt at a)
Data:

Fw:75kg
g:-9.80m/s^2
m: 500N
a=?

Formula:
F=ma
a=F/m
a=500N/75kg
a=6.66m/s^2

My attempt at b)
Formula:

a=F/m
a=1500N/75kg
a=20m/s^2

My attempt at c)
Formula

a=f/m
a=75N/75kg
a=1 m/s^2
first of all I want to thank anyone that gives their time and effort to help me do this homework that is due very soon. I donâ€™t think I understand it very well and I am start to get really worried. Secondly, if you notice any problems, can you please show me how to do it? Id appreciate any help I can get. Thanks a ton people.

2. Oct 26, 2009

### willem2

Q1: ok

Q2; you didn't copy question A. F=19.6 N is wrong however (and F=2.0kg is wronger)
Let T be the tension in the rope, and then find the net force on both blocks
from gravity and this unknown tension. Then use F=ma for both blocks, and the fact
that a is the same for both blocks to find T.

Q3-a: The answer is OK. You should have been using 2as=Vf^2-Vi^2 and then a would have
become negative by itself. The sign of a also depends on what direction is chosen as
the positive direction. Here it's the direction that the car drives, but that isn't always the case.
Q3-b: 4.9 m/s^2 isn't a force and neither is 1200 kg. What are the forces?
Q4: I don't do slugs, but you need to find the net force, wich includes the friction force that is opposite to the pulling force.
Q5: You need to find the net force here as well. In physics weight is a force, the force of gravity, and it is equal to the mass times the acceleration of gravity mg with g = 9.8 m/s^2
so the weight of the person is 735N and not 75 kg.