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Date Submitted: 11/14/2014 12:39 PM
Chapter 5 pg 143
1. Mass 6.00 kg moved by 18.0N. What is its acceleration?
F=ma Divide both sides by m
a=Fm
a=18N/6.00kg convert N = 18N * 1kg*m/s^2 = 18kg*m/s^2
a=18kg*m/s^2/ 6.00 kg = 3m/s^2
a=3m/s^2
7. Truck pulls trailer with friction force of 870 N and a coefficient of friction 0.23. What is the trailer’s normal force.
Ff=μFN
870N=.023(FN) divide by .023
FN=870N/.023
FN=37826.7N
9. mass 13.0 kg dropped from a cliff. Find its weight
Fw=mg
Fw=13.0kg * 9.8m/s^2
Fw=127.4kg*m/s^2
Fw=127.4N
13. A force of 175N needed to keep a 640N engine from sliding. What is the coefficient of static friction?
Ff=μFN
Ff=175N
FN=640N
175N=μ640N divide by 640N
µ=175N/640N
µ=.273
1400Kg
1400Kg
2100N-425N=1675N
a=F/m
a=1675N/1400kg=1675kg*m/s^2/1400kg=1.2m/s^2
2100N-425N=1675N
a=F/m
a=1675N/1400kg=1675kg*m/s^2/1400kg=1.2m/s^2
15. find acceleration of a forklift mass 1400kg pushed by force 2100N opposed by frictional force of 425N.
2100N
2100N
a = ??
a = ??
425N
425N
Chapter 6 page 163
3. Box pushed with a force 125 N for 2.00 min. What is the impulse?
impulse=Ft
Impulse=125N*120s
Impulse=15000Ns=15000(kg*m/s^2)(s)
Impusle=15000kg*m/s
7. 15.0g bullet fired. Muzzle velocity 3250m/s from a rifle with mass 4.74kg and barrel length 75.0 cm.
A) How long is bullet in barrel? d/s=t
B) what is the force on the bullet while in the barrel? Ft=mvf-mvi
C) Find the bullet’s momentum as it leaves barrel. p=mv
a) t=75cm/3250m/s 100cm=1m 75cm/1* 1m/100cm= .75m
t=.75m/3250m/s=2.307E-4s
time in barrel is 2.307E-4 seconds
b) F=mvf-mvit
F=15.0g3250ms-15.0g0ms2.307X10-4
F=15.0g*3250ms/2.307X10-4s
F=48750g*m/s/2.307X10-4s=48.750kg*m/s/2.307X10-4s
F=211313.394kg*m/s2 X 1N/1kg*m/s2=211313.394N
211,313.394N of force on the bullet as it leaves the barrel.
c) p=mv
p=15.0g * 3250m/s
p=48.75kg*m/s
m1v1+m2v2=m1v’1+m2v’2
(575g*3.05m/s)+(425g*0m/s)=(575g*v’1)+(425g*4.03m/s)...