(a) A weight of mass 2.10 kg slides down a vertical, frictionless chute onto a v
ID: 1772050 • Letter: #
Question
(a) A weight of mass 2.10 kg slides down a vertical, frictionless chute onto a vertical spring with force constant 533 N/m. If the weight was initially 171 cm above the top of the spring, how much does the weight compress the spring before coming momentarily to rest? (b) For the motion described in Part (a), by how much is the spring compressed at the instant that the net force on the weight is zero? (c) Same as above except that friction is present and the maximum compression of the spring is fraction 0.430 of what was found in Part (a). How much work was done by the force of friction while the weight was sliding down and compressing the spring?
a) _____ cm
b) _____ cm
c) _____ J
Explanation / Answer
(a) We find velocity of the mass as it just touches the spring ( just before compression starts).
h=1.71 m, g=9.8m/s2, u=0;
v2-u2=2*g*h, v2=2*9.8*1.71
v2=33.516 (m/s)2
Now the kinetic energy of the mass before compression will be completely changed into the potential energy of the spring after compression.
0.5*m*v2=0.5*k*x2 {0.5*k*x2 is the potential energy stored in spring with k spring constant and x compression}
0.5*2.1*33.516=0.5*533*x2
x=0.3633 m
(b) downward force on the mass=m*g
upward force on the mass= k*x
net force =0 that is,
m*g=k*x
2.1*9.8=533.x
x=0.0386 m
(c) New compression value is x'= 0.0386*0.430=0.0166 m
since friction is also there,
net force =0 when
m*g=(k*x)+f {here f is the frictional force}
2.1*9.8=533*0.0166+f
f=11.74 Newton
total displacement of the mass = 1.71 m(sliding distance)+ 0.0166 m (compression)= 1.7266 m
Work Done = force*displacement
W.D=11.74*1.7266
W.D =20.270 joules
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