Question
A flatbed truck with a box sitting on the bed accelerates from rest on level ground. The truck’s speed as a
function of time is 15t –2t^2 during this acceleration. The truck reaches its maximum speed and cruises along at this speed for 4 minutes. It then slows to a stop with an acceleration of 2 m/s/s.
a) What is the net force as a function of time, acting on the truck during the speeding up portion?
b) What is the maximum speed the truck attains?
c) At what time does the truck reach maximum speed?
d) How far does the truck travel during the entire trip?
e) How much work is done on the 588 N box in the bed of the truck while speeding up?
f) What force does that work on the box?
g) How much total work is done on the box during the entire trip?
function of time is 15t –2t^2 during this acceleration. The truck reaches its maximum speed and cruises along at this speed for 4 minutes. It then slows to a stop with an acceleration of 2 m/s/s.
a) What is the net force as a function of time, acting on the truck during the speeding up portion?
b) What is the maximum speed the truck attains?
c) At what time does the truck reach maximum speed?
d) How far does the truck travel during the entire trip?
e) How much work is done on the 588 N box in the bed of the truck while speeding up?
f) What force does that work on the box?
g) How much total work is done on the box during the entire trip?
Answers
v=15t-2t²
a=dv/dt = 15 - 4t.
(a)F=m•a=m• (15-4t),
For finding v(max):
a=15 - 4t=0
15=4t
(c) t=15/4 (s)
(b)v(max)=15t-2t² =
=15•15/4 -2•(15/4)²=28.125 m/s.
(d) s1=∫(15t-2t²)dt=15t²/2 – 2t³/3 =
=15•15²/2•4² -2•15³/3•4³ = 70.31 m.
s2 = v(max) •t=28.125•4•60=6750 m.
s3= v²/2a=(28.125)²/2•2 = 197.8 m
s1+s2+s3 = 70.31+6750+197.8 =7018 m.
(e) The work of the net force during the sppeding up portion
W1=ΔKE=mv²/2=(588/9.8) •28.125²/2=2.37•10^4 J.
(f) F=m•a=m• (15-4t),
(g) W2=0 (since the net force is zero),
W3=ΔKE =0- mv²/2=0-(588/9.8) •28.125²/2=
=0-2.37•10^4 J= - 2.37•10^4 J.
W=W1+W2+W3=0
a=dv/dt = 15 - 4t.
(a)F=m•a=m• (15-4t),
For finding v(max):
a=15 - 4t=0
15=4t
(c) t=15/4 (s)
(b)v(max)=15t-2t² =
=15•15/4 -2•(15/4)²=28.125 m/s.
(d) s1=∫(15t-2t²)dt=15t²/2 – 2t³/3 =
=15•15²/2•4² -2•15³/3•4³ = 70.31 m.
s2 = v(max) •t=28.125•4•60=6750 m.
s3= v²/2a=(28.125)²/2•2 = 197.8 m
s1+s2+s3 = 70.31+6750+197.8 =7018 m.
(e) The work of the net force during the sppeding up portion
W1=ΔKE=mv²/2=(588/9.8) •28.125²/2=2.37•10^4 J.
(f) F=m•a=m• (15-4t),
(g) W2=0 (since the net force is zero),
W3=ΔKE =0- mv²/2=0-(588/9.8) •28.125²/2=
=0-2.37•10^4 J= - 2.37•10^4 J.
W=W1+W2+W3=0