Question
A track consists of a frictionless incline plane, which is a height of 0.5m, and a rough horizontal section with a coefficient of kinetic friction 0.02. Block A, whose mass is 1.5kg, is released from the top of the incline plane, slides down and collides instantaneously and inelastically with identical block B at the lowest point. The two blocks move to the right through the rough section of the track until they stop.
a. Determine the initial potential energy of block A.
b. Determine the kinetic energy of block A at the lowest point, just before the collision.
c. Find the speed of the two blocks just after the collision.
d. Find the kinetic energy of the two blocks just after the collision.
e. How far will the two blocks travel on the rough section of the track?
f. How much work will the friction force do during this time?
a. Determine the initial potential energy of block A.
b. Determine the kinetic energy of block A at the lowest point, just before the collision.
c. Find the speed of the two blocks just after the collision.
d. Find the kinetic energy of the two blocks just after the collision.
e. How far will the two blocks travel on the rough section of the track?
f. How much work will the friction force do during this time?
Answers
a. PE = mgh=1.5•9.8•0.5 = 7.35 J
b. KE=PE = 7.35 J
c. KE = mv²/2
v= sqrt(2•KE/m) =
=sqrt(2•7.35/1.5)=3.13 m/s
mv + 0 = 2mu
u=mv/2m = v/2= 3.13/2 =1.565 m/s
d. KE₂=2m•u²/2 = 2•1.5•1.565²/2 =3.68 J
e. KE₂ = W(fr)=F(fr)s =μ•N•s
= μ•2mg•s.
s= KE₂/μ•2mg=3.68/0.02•2•1.5•9.8 =6.26 m
f. W(fr) = KE₂ = 3.68 J
b. KE=PE = 7.35 J
c. KE = mv²/2
v= sqrt(2•KE/m) =
=sqrt(2•7.35/1.5)=3.13 m/s
mv + 0 = 2mu
u=mv/2m = v/2= 3.13/2 =1.565 m/s
d. KE₂=2m•u²/2 = 2•1.5•1.565²/2 =3.68 J
e. KE₂ = W(fr)=F(fr)s =μ•N•s
= μ•2mg•s.
s= KE₂/μ•2mg=3.68/0.02•2•1.5•9.8 =6.26 m
f. W(fr) = KE₂ = 3.68 J